Three Dimensional Geometry: JEE Main 2026 PYQs

Answer: Option (3)

Key idea: Intersection of lines in 3D space and condition for point on xy-plane

1. Parametrize both lines: Line 1 with parameter $t$ gives $(3t-1, 5t-a, 7t-b-1)$ and Line 2 with parameter $s$ gives $(s+2, 4s+b, 7s+2a)$
2. Equate coordinates at intersection: From $x$: $3t - s = 3$; From $y$: $5t - 4s = a + b$; From $z$: $7(t-s) = 2a + b + 1$
3. Since intersection lies on $xy$-plane, $z = 0$. Using Line 1: $7t - b - 1 = 0 \Rightarrow t = \frac{b+1}{7}$
4. Solve for $s$ from equation 1: $s = 3t - 3$. Substitute into the $y$ equation to get $-7t + 12 = a + b$
5. Substitute $t = \frac{b+1}{7}$ into $-7t + 12 = a + b$ to get $11 = a + 2b$
6. Use the $z$ equation to get another relation: $2a = 18 - 3b$. Solving both equations gives $a = 3, b = 4$
7. Therefore, $a + b = 3 + 4 = 7$

Answer: Option (3)

Key idea: Foot of perpendicular from a point to a line in 3D, Cross product for direction vector

1. Find direction vector of line $L$ using cross product: $\vec{d} = (2\hat{i} + 2\hat{j} + \hat{k}) \times (\hat{i} + 2\hat{j} + 2\hat{k}) = 2\hat{i} - 3\hat{j} + 2\hat{k}$
2. Write equation of line $L$ passing through $(1,1,1)$: $\vec{r} = (1,1,1) + t(2,-3,2)$
3. Point $P$ on line $L$ is $(1+2t, 1-3t, 1+2t)$. Since $\vec{OP} \perp \vec{d}$, we have $\vec{OP} \cdot \vec{d} = 0$
4. Solve: $2(1+2t) - 3(1-3t) + 2(1+2t) = 0 \Rightarrow 1 + 17t = 0 \Rightarrow t = -\frac{1}{17}$
5. Find $P$: $a = \frac{15}{17}$, $b = \frac{20}{17}$, $c = \frac{15}{17}$, so $a+b+c = \frac{50}{17}$
6. Calculate: $34(a+b+c) = 34 \times \frac{50}{17} = 100$

Answer: Option (1)

Key idea: 3D Geometry: Foot of perpendicular from a point to a line, midpoint formula, and perpendicularity condition

1. Use midpoint formula: If $M(0, \frac{3}{4}, \frac{-1}{4})$ is midpoint of $PA$, then $A = (-a, \frac{3}{2} - b, \frac{-1}{2})$
2. Since $A$ lies on the line, write $A = (1+2t, 2+t, \alpha+3t)$ and equate: $1+2t = -a$, $2+t = \frac{3}{2}-b$, $\alpha+3t = \frac{-1}{2}$
3. From the second equation: $t = -\frac{1}{2} - b$. Substitute in first: $a = 2b$. Substitute in third: $\alpha = 1 + 3b$
4. Apply perpendicularity: $\vec{PA} \cdot \vec{d} = 0$ where $\vec{PA} = (-2a, \frac{3}{2}-2b, \frac{-1}{2})$ and $\vec{d} = (2, 1, 3)$
5. Simplify: $-4a - 2b = 0 \Rightarrow a = -\frac{b}{2}$. Combined with $a = 2b$ gives $b = 0$, hence $a = 0$ and $\alpha = 1$
6. Calculate: $a^2 + b^2 + \alpha^2 = 0^2 + 0^2 + 1^2 = 1$

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Answer: Option (2)

Key idea: Distance between point and line along a given direction; Intersection of two lines in 3D space

1. Verify that point $P(-2, -8, 6)$ lies on line $L_2: \frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ by checking all ratios equal 3
2. Write parametric equations: $L_1: (1+s, 1+2s, -s)$ and $L_2: (-5+t, -5-t, 2t)$
3. Find intersection of $L_1$ and $L_2$ by equating coordinates: $t-s=6$, $-t-2s=6$, $2t+s=0$
4. Solve the system to get $t=2$ and $s=-4$, giving intersection point $Q(-3, -7, 4)$
5. Calculate distance squared: $d^2 = ((-2)-(-3))^2 + ((-8)-(-7))^2 + (6-4)^2 = 1 + 1 + 4 = 6$

Answer: Option (2)

Key idea: Shortest distance between skew lines using vector formula

1. Identify $\vec{a_1} = \frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}$, $\vec{b_1} = 2\hat{i} - 5\hat{j} + 6\hat{k}$, $\vec{a_2} = -\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}$, $\vec{b_2} = \hat{j} - \hat{k}$
2. Calculate $\vec{a_2} - \vec{a_1} = -\hat{i} - 2\hat{j} - 3\hat{k}$
3. Find cross product $\vec{b_1} \times \vec{b_2} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and its magnitude $|\vec{b_1} \times \vec{b_2}| = 3$
4. Calculate dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 1 - 4 - 6 = -9$
5. Apply formula: $d = \frac{|{-9}|}{3} = 3$

Answer: Option (1)

Key idea: Image of a point with respect to a line in 3D geometry

1. Find midpoint $M = \left(\frac{3a+1}{2}, \frac{a^2-a}{2}, \frac{a+1}{4}\right)$ of $P(a, 2a, 1)$ and $Q(2a+1, a^2-3a, \frac{a-1}{2})$
2. Since $M$ lies on line $\frac{x-2}{3} = \frac{y-1}{2} = \frac{z}{1} = t$, we get $t = \frac{a+1}{4}$
3. Substitute in line equations: $\frac{3a+1}{2} = 3t+2$ and $\frac{a^2-a}{2} = 2t+1$
4. From first equation with $t = \frac{a+1}{4}$: $\frac{3a+1}{2} = \frac{3a+11}{4}$ gives $a = 3$
5. Verify perpendicularity: $\vec{PQ} \cdot \vec{d} = 0$ gives $4a^2 - 13a + 3 = 0$, so $a = 3$ or $a = \frac{1}{4}$
6. Check both values in midpoint condition: only $a = 3$ satisfies all conditions

Answer: Option (3)

Key idea: Distance from a point to a line in 3D using cross product formula

1. Identify point on line $A(2, 5, 2)$ and direction vector $\vec{b} = (2, 3, 4)$
2. Find vector $\overrightarrow{AP} = (5-2, 6-5, 7-2) = (3, 1, 5)$
3. Calculate cross product: $\overrightarrow{AP} \times \vec{b} = (-11, -2, 7)$
4. Find $|\overrightarrow{AP} \times \vec{b}|^2 = 121 + 4 + 49 = 174$ and $|\vec{b}|^2 = 4 + 9 + 16 = 29$
5. Apply formula: $d^2 = \frac{|\overrightarrow{AP} \times \vec{b}|^2}{|\vec{b}|^2} = \frac{174}{29} = 6$

Answer: 17

Key idea: Intersection of lines in 3D space and distance from origin

1. Write both lines in parametric form: Line 1 gives $x = 1 + a\lambda$, $y = 1 - \lambda$, $z = -1$ and Line 2 gives $x = 4 + 2\mu$, $y = 0$, $z = -1 + a\mu$
2. Equate $y$-coordinates: $1 - \lambda = 0 \Rightarrow \lambda = 1$
3. Equate $z$-coordinates: $-1 = -1 + a\mu \Rightarrow a\mu = 0$. Since $a \neq 0$, we get $\mu = 0$
4. Equate $x$-coordinates: $1 + a(1) = 4 + 2(0) \Rightarrow a = 3$
5. Substitute $\lambda = 1$ in Line 1 to get intersection point: $(4, 0, -1)$
6. Calculate square of distance from origin: $d^2 = 4^2 + 0^2 + (-1)^2 = 16 + 0 + 1 = 17$

Answer: Option (3)

Key idea: Finding points on a line at given distance from a point, and calculating centroid of a triangle in 3D

1. Write the parametric form of the line: $(8t-3, 2t+4, 2t-1)$ where $t$ is the parameter
2. Use distance formula: Points at distance 6 from $R(1,2,3)$ satisfy $(8t-4)^2 + (2t+2)^2 + (2t-4)^2 = 36$
3. Simplify to get $72t^2 - 72t + 36 = 36$, which gives $2t(t-1) = 0$, so $t = 0$ or $t = 1$
4. Find coordinates: $P(-3, 4, -1)$ at $t=0$ and $Q(5, 6, 1)$ at $t=1$
5. Calculate centroid: $(\alpha, \beta, \gamma) = \left(\frac{-3+5+1}{3}, \frac{4+6+2}{3}, \frac{-1+1+3}{3}\right) = (1, 4, 1)$
6. Therefore $\alpha + \beta + \gamma = 1 + 4 + 1 = 6$

Answer: Option (3)

Key idea: Image of a point with respect to a line in 3D space

1. Find the foot of perpendicular $M$ from point $(1, 2, 7)$ to the line $\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ by setting $\vec{PM} \perp$ direction vector $(1, 1, 2)$
2. Parametrize the line as $(t, 1+t, 2+2t)$ and solve $(t-1, t-1, 2t-5) \cdot (1, 1, 2) = 0$ to get $t = 2$, so $M = (2, 3, 6)$
3. Find the image $P'$ using $M$ as midpoint: $P' = 2M - P = (3, 4, 5)$
4. Calculate distance from $(a, 2, 5)$ to $P'(3, 4, 5)$: $\sqrt{(a-3)^2 + 4} = 4$
5. Solve $(a-3)^2 = 12$ to get $a = 3 \pm 2\sqrt{3}$
6. Sum of all possible values: $(3 + 2\sqrt{3}) + (3 - 2\sqrt{3}) = 6$

Answer: Option (3)

Key idea: Image of a point in a line, foot of perpendicular, distance formula in 3D

1. Step 1: Find foot of perpendicular $F$ from $P(1,6,a)$ to line $L$. Since $F$ is midpoint of $P$ and its image $Q(\frac{a}{3}, 0, a+c)$, we get $F = (\frac{3+a}{6}, 3, \frac{2a+c}{2})$.
2. Step 2: Since $F$ is on line $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$, from $y$-coordinate: $1+2t=3 \Rightarrow t=1$. This gives $F=(1, 3, a-1+b)$.
3. Step 3: From $x$-coordinate: $1 = \frac{3+a}{6} \Rightarrow a=3$. From $z$-coordinate and perpendicularity condition $t = \frac{11+b}{5+b^2} = 1$, we get $b^2-b-6=0 \Rightarrow b=3$ (since $b>0$).
4. Step 4: Using $z$-coordinate equation: $2+b = \frac{6+c}{2} \Rightarrow c=4$. So $F=(1,3,5)$ and line $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.
5. Step 5: Point on $L$ is $(\lambda, 1+2\lambda, 2+3\lambda)$. Distance from $F$ to $S$ is $|SF| = \sqrt{14}|\lambda-1| = 2\sqrt{14}$, giving $|\lambda-1|=2$.
6. Step 6: Since $\alpha > 0$, we take $\lambda=3$ (not $-1$). Thus $S=(3, 7, 11)$ and $\alpha+\beta+\gamma = 3+7+11=21$.

Answer: Option (2)

Key idea: Cross product to find perpendicular line, angle between lines using dot product

1. Find direction vector of $L$ using cross product: $\vec{d_L} = \vec{d_1} \times \vec{d_2}$ where $\vec{d_1} = (3, 5, 7)$ and $\vec{d_2} = (1, 4, 7)$
2. Calculate: $\vec{d_L} = (7, -14, 7) = 7(1, -2, 1)$, so direction ratios are $(1, -2, 1)$
3. For angle with $L_3$ having direction $(2, 1, 2)$: $\cos \theta = \frac{|\vec{d_L} \cdot \vec{d_{L_3}}|}{|\vec{d_L}| \cdot |\vec{d_{L_3}}|} = \frac{|2|}{\sqrt{6} \cdot 3} = \frac{2}{3\sqrt{6}}$
4. Find $\sin \theta$: $\sin^2 \theta = 1 - \frac{4}{54} = \frac{25}{27}$, so $\sin \theta = \frac{5}{3\sqrt{3}}$
5. Calculate $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{5}{3\sqrt{3}} \cdot \frac{3\sqrt{6}}{2} = \frac{5\sqrt{2}}{2}$

Answer: Option (3)

Key idea: Shortest distance between two skew lines using the formula d = |((P₂ - P₁) · (b₁ × b₂))| / |b₁ × b₂|

1. Identify points and direction vectors: $P_1 = (4, 3, 2)$, $\vec{b_1} = (1, 2, -3)$ and $P_2 = (-2, 6, 5)$, $\vec{b_2} = (2, 4, -5)$
2. Calculate $P_2 - P_1 = (-6, 3, 3)$
3. Find cross product: $\vec{b_1} \times \vec{b_2} = (2, -1, 0)$
4. Calculate magnitude: $|\vec{b_1} \times \vec{b_2}| = \sqrt{4 + 1 + 0} = \sqrt{5}$
5. Find dot product: $(P_2 - P_1) \cdot (\vec{b_1} \times \vec{b_2}) = (-6)(2) + (3)(-1) + (3)(0) = -15$
6. Apply formula: $d = \frac{|-15|}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$

Answer: 162

Key idea: Image of a point in a line, Area of parallelogram using cross product

1. Find image $R$ of $P(0,-5,0)$ in line $\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}$: Foot of perpendicular $M=(-1,-1,1)$, so $R=2M-P=(-2,3,2)$
2. Find image $S$ of $Q(0,-\frac{1}{2},0)$ in line $\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}$: Foot of perpendicular $N=(-1,-1,1)$, so $S=2N-Q=(-2,-\frac{3}{2},2)$
3. Verify $PQRS$ is a parallelogram: Check $\vec{PQ}=\vec{SR}=(0,\frac{9}{2},0)$ and $\vec{PS}=\vec{QR}=(-2,\frac{7}{2},2)$
4. Calculate area using cross product: $\vec{PQ}\times\vec{PS}=(9,0,9)$
5. Find magnitude: $|\vec{PQ}\times\vec{PS}|=\sqrt{81+0+81}=9\sqrt{2}$
6. Square of area: $(9\sqrt{2})^2=81\times 2=162$

Answer: $\frac{\sqrt{146}}{7}$

Key idea: Distance between parallel lines in 3D, foot of perpendicular from a point to a line

1. Find the foot of perpendicular $(1, \mu, 2)$ on the line by substituting in parametric form: $t = -3$ gives $\mu = 3$
2. Use perpendicularity condition: vector from $(1, 3, 2)$ to $(\lambda, 2, 3)$ is perpendicular to direction $(1, 2, 1)$, giving $(\lambda-1) - 2 + 1 = 0$, so $\lambda = 2$
3. Identify the two lines with $\lambda = 2, \mu = 3$. Both have direction vector $(2, 3, 6)$, so they are parallel
4. For parallel lines, use distance formula: $d = \frac{|\vec{P_1P_2} \times \vec{d}|}{|\vec{d}|}$ where $\vec{P_1P_2} = (1, 1, -1)$
5. Calculate cross product: $(1, 1, -1) \times (2, 3, 6) = (9, -8, 1)$ with magnitude $\sqrt{146}$
6. Distance = $\frac{\sqrt{146}}{7}$

Answer: 4

Key idea: 3D Geometry: Line perpendicular to two lines, intersection of lines, distance formula

1. Find direction vectors: $\vec{b_2} = (1, 2, 2)$ for $L_2$ and $\vec{b_3} = (2, 2, 1)$ for $L_3$
2. Direction of $L_1$ perpendicular to both: $\vec{b_1} = \vec{b_2} \times \vec{b_3} = (-2, 3, -2)$
3. Since $L_1$ passes through origin: $L_1: \vec{r} = \lambda(-2, 3, -2)$
4. Find intersection of $L_1$ and $L_2$ by solving $(-2\lambda, 3\lambda, -2\lambda) = (3+t, 2t-1, 2t+4)$, giving $\lambda = -1, t = -1$, so point $P = (2, -3, 2)$
5. Point on $L_3$ is $(3+2s, 3+2s, 2+s)$. Distance from $P$ equals $\sqrt{17}$ gives $9s^2 + 28s + 20 = 0$
6. Solving: $s = -2$ or $s = -\frac{10}{9}$. Since $a \in \mathbb{Z}$, use $s = -2$ giving $(-1, -1, 0)$
7. Therefore $(a+b+c)^2 = (-1-1+0)^2 = 4$