QJEEM24Maddab: JEE Main 2024 Mathematics - Class 12 JEEM24Maddab Solution: The shortest distance between the lines and is:

Given lines are

$\frac{x-3}{4}=\frac{y-(-7)}{-11}=\frac{z-1}{5}$ and

$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z-(-2)}{1}$

Shortest distance between two lines,

$\begin{aligned}& d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|} \\& \vec{a}_2-\vec{a}_1=2 \hat{i}+16 \hat{j}-3 \hat{k} \text { and } \\& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\4 & -11 & 5 \\3 & -6 & 1\end{array}\right|=19 \hat{i}+11 \hat{j}+9 \hat{k} \\& \therefore \quad d=\frac{187}{\sqrt{563}}\end{aligned}$