JEE Main 2024 — 30 January, Shift 1 — Question 22

$\mathrm{L}_{1}: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}_{2}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}}$

$\mathrm{d}_{1}=$ shortest distance between $\mathrm{L}_{1} \& \mathrm{~L}_{2}$

$=\left|\frac{\left(\overrightarrow{\mathrm{a}}_{2}-\overrightarrow{\mathrm{a}}_{1}\right) \cdot\left(\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}}_{2}\right)}{\left|\left(\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}}_{2}\right)\right|}\right|$

$\mathrm{d}_{1}=2$

$\mathrm{L}_{3}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+8}{-7}=\frac{\mathrm{z}-4}{5}, \mathrm{~L}_{4}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-6}{-3}$

$\mathrm{d}_{2}=$ shortest distance between $\mathrm{L}_{3} \& \mathrm{~L}_{4}$ $\mathrm{d}_{2}=\frac{12}{\sqrt{3}}$

Hence $=\frac{32 \sqrt{3} \mathrm{~d}_{1}}{\mathrm{~d}_{2}}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16$