QJEEM24Mafe3a: JEE Main 2024 Mathematics - Class 12 JEEM24Mafe3a Solution: If the shortest distance between the lines and is , then the…

$\begin{aligned}& \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \\& \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}\end{aligned}$

the shortest distance between the lines

$=\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot\left(\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right)}{\left|\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right|}\right|$

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

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

$\begin{aligned}& \frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right| \\& 3=|\lambda-4| \\& \lambda-4= \pm 3 \\& \lambda=7,1\end{aligned}$

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