Q40 / 75 · JEE Main 2026 (22 January Shift 1) · Mathematics
Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ 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
- Correct answerA$4 \sqrt{\frac{7}{5}}$
- B$2 \sqrt{\frac{7}{4}}$
- C$7 \sqrt{\frac{5}{4}}$
- D$4 \sqrt{\frac{5}{7}}$
Solution
Let $P(2\lambda + 1, -3\lambda - 1, \lambda)$
Then $4\lambda^2 + 9\lambda^2 + \lambda^2 = 16 \cdot 14 \Rightarrow \lambda = \pm 4 \Rightarrow -4$
(nearer to origin)
$\therefore P(-7, 11, -4)$
$\therefore$ Shortest distance $= \dfrac{\begin{vmatrix} 2 & -1 & 7 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix}}{\left| \hat{i} \quad \hat{j} \quad k \right| \begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix}}$
$= \dfrac{28}{\sqrt{1 + 25 + 9}} = \dfrac{4\sqrt{7}}{\sqrt{5}}$
Then $4\lambda^2 + 9\lambda^2 + \lambda^2 = 16 \cdot 14 \Rightarrow \lambda = \pm 4 \Rightarrow -4$
(nearer to origin)
$\therefore P(-7, 11, -4)$
$\therefore$ Shortest distance $= \dfrac{\begin{vmatrix} 2 & -1 & 7 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix}}{\left| \hat{i} \quad \hat{j} \quad k \right| \begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix}}$
$= \dfrac{28}{\sqrt{1 + 25 + 9}} = \dfrac{4\sqrt{7}}{\sqrt{5}}$
Hence, the correct answer is option: 'A'.