Three Dimensional Geometry – Lines – JEE Main 22 Jan 2026 Shift 1

Solution:

Any point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z}{1}$ can be written as:
$$ P(2\lambda+1, -3\lambda-1, \lambda) $$
The distance of $P$ from point $A(1, -1, 0)$ is $4\sqrt{14}$.
$$ AP^2 = (2\lambda)^2 + (-3\lambda)^2 + (\lambda)^2 = 4\lambda^2 + 9\lambda^2 + \lambda^2 = 14\lambda^2 $$

Given $AP = 4\sqrt{14} \Rightarrow AP^2 = 16 \times 14$.
$$ 14\lambda^2 = 16 \times 14 \Rightarrow \lambda^2 = 16 \Rightarrow \lambda = \pm 4 $$

Case 1: $\lambda = 4$
$P_1(9, -13, 4)$. Distance from origin $OP_1 = \sqrt{81 + 169 + 16} = \sqrt{266}$.

Case 2: $\lambda = -4$
$P_2(-7, 11, -4)$. Distance from origin $OP_2 = \sqrt{49 + 121 + 16} = \sqrt{186}$.

Since $P$ is nearer to the origin, we choose $P(-7, 11, -4)$.
So, $\alpha = -7, \beta = 11, \gamma = -4$.

The two lines for shortest distance are:
$L_1: \frac{x+7}{1}=\frac{y-11}{2}=\frac{z+4}{3}$ passing through $A(-7, 11, -4)$ with $\vec{b_1} = \hat{i} + 2\hat{j} + 3\hat{k}$.
$L_2: \frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$ passing through $B(-5, 10, 3)$ with $\vec{b_2} = 2\hat{i} + \hat{j} + \hat{k}$.

Vector $\vec{AB} = (-5 – (-7))\hat{i} + (10 – 11)\hat{j} + (3 – (-4))\hat{k} = 2\hat{i} – \hat{j} + 7\hat{k}$.

Cross product $\vec{b_1} \times \vec{b_2}$:
$$ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix} $$
$$ = \hat{i}(2-3) – \hat{j}(1-6) + \hat{k}(1-4) = -\hat{i} + 5\hat{j} – 3\hat{k} $$
Magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-1)^2 + 5^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$.

Shortest Distance (SD) formula:
$$ SD = \left| \frac{\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| $$
$$ \vec{AB} \cdot (\vec{b_1} \times \vec{b_2}) = (2)(-1) + (-1)(5) + (7)(-3) = -2 – 5 – 21 = -28 $$

$$ SD = \frac{|-28|}{\sqrt{35}} = \frac{28}{\sqrt{35}} $$
$$ SD = \frac{28}{\sqrt{5} \sqrt{7}} = \frac{4 \times 7}{\sqrt{5} \sqrt{7}} = \frac{4\sqrt{7}}{\sqrt{5}} = 4\sqrt{\frac{7}{5}} $$

Ans. (2)