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Question 19 — JEE Main 22 Jan 2026 (S I)

Question

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the point on the line $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{-3}=\mathrm{z}$ at a distance $4 \sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance,
between the lines $\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$ and $\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$, is equal to

[JEE MAIN 22 Jan 2026 S I]

(1) $7 \sqrt{\frac{5}{4}}$
(2) $4 \sqrt{\frac{7}{5}}$
(3) $4 \sqrt{\frac{5}{7}}$
(4) $2 \sqrt{\frac{7}{4}}$

Solution

line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z}{1}=\lambda$
$\mathrm{P}(2 \lambda+1,-3 \lambda-1, \lambda) \cdot \mathrm{A}(1,-1,0)$
$\mathrm{AP}=\sqrt{4 \lambda^{2}+9 \lambda^{2}+\lambda^{2}}=4 \sqrt{14}$
$\Rightarrow 14 \lambda^{2}=16 \times 14 \Rightarrow \lambda= \pm 4$
$\lambda=4 \mathrm{P}(9,-13,4) ; \mathrm{OP}=\sqrt{81+169+16}=\sqrt{266}$ greatest
$\lambda=-4 \mathrm{P}(-7,11,-4) ; \mathrm{OP}=\sqrt{49+121+16}=\sqrt{186}$ least
$\mathrm{P}(-7,11,-4) \equiv(\alpha, \beta, \gamma) ; \alpha=-7, \beta=11, \gamma=-4$
Shortest distance bw skew lines $=\frac{\left|\begin{array}{ccc}\mathrm{x}_{2}-\mathrm{x}_{1} & \mathrm{y}_{2}-\mathrm{y}_{1} & \mathrm{z}_{2}-\mathrm{z}_{1} \\ \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2}\end{array}\right|}{\left\|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2}\end{array}\right\|}$

=\frac{\left|\begin{array}{ccc} -2 & 1 & -7 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{array}\right|}{\left\|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{array}\right\|}=\frac{|-2(-1)+5-7(-3)|}{|-\hat{i}+5 \hat{j}-3 \hat{k}|}=\frac{28}{\sqrt{35}}=4 \sqrt{\frac{7}{5}} \text { Answer }(2)