JEE Main 2024 — 27 January, Shift 1 — Question 8

$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$

$\frac{\mathrm{x}-\lambda}{2}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}-2}{-5}$

the shortest distance between the lines $=\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot\left(\overrightarrow{\mathrm{d}_{1}} \times \overrightarrow{\mathrm{d}_{2}}\right)}{\left|\overrightarrow{\mathrm{d}_{1}} \times \overrightarrow{\mathrm{d}_{2}}\right|}\right|$

$=\left|\frac{\left|\begin{array}{ccc}\lambda-4 & 0 & 2\\ 1 & 2 & -3 \\2 & 4 & -5\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k}\\ 1 & 2 & -3 \\2 & 4 & -5\end{array}\right|}\right|$

$=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right|$

$\frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right|$

$3=|\lambda-4|$

$\lambda-4= \pm 3$

$\lambda=7,1$

Sum of all possible values of $\lambda$ is $=8$