QJEEM24Ma1863: JEE Main 2024 Mathematics - Class 12 JEEM24Ma1863 Solution: If is the shortest distance between the lines and is the…

$\mathrm{L}_1: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}_2: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}}$

$\mathrm{d}_1=$ shortest distance between $\mathrm{L}_1 ~\& \mathrm{~L}_2$

$\begin{aligned}& =\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}\right| \\& d_1=2 \\& \mathrm{~L}_3: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+8}{-7}=\frac{\mathrm{z}-4}{5}, \mathrm{~L}_4: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-6}{-3} \\& \mathrm{~d}_2=\text { shortest distance between } \mathrm{L}_3 ~\& \mathrm{~L}_4 \\& \mathrm{~d}_2=\frac{12}{\sqrt{3}} \text { Hence } \\& =\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16\end{aligned}$