For what value of k are the lines $\frac{x}{k}=2-y=\frac{z-1}{2}\text{ and }\frac{x+1}{3}=\frac{y-1}{k}=z+2$ perpendicular to each other ?

Two lines are perpendicular to each other if product of their direction cosines or direction ratios is zero

Straight lines $L_1$ and $L_2$ are ${\mathrm{L}}_1:\frac{\mathrm{x}}{\mathrm{k}}$$=2-y=\frac{z-1}{2}$

${\mathrm{L}}_2:\frac{\mathrm{x}+1}{3}=\frac{\mathrm{y}-1}{\mathrm{k}}=\mathrm{z}+2$

It can be rewritten as

$\begin{array}{r}{\mathrm{L}}_1:\frac{\mathrm{x}+0}{\mathrm{k}}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-2}{2}\\ {\mathrm{L}}_2:\frac{\mathrm{x}+1}{3}=\frac{\mathrm{y}-1}{\mathrm{k}}=\frac{\mathrm{z}+2}{1}\end{array}$

Direction ratios of line  ${\mathrm{L}}_1:(\mathrm{k},-1,2)$

Direction ratios of line  ${\mathrm{L}}_2:(3,\mathrm{k},1)$

So, $3\mathrm{k}+\mathrm{k}(-1)+2=0$

$\begin{array}{r}\Rightarrow 3\mathrm{k}-\mathrm{k}+2=0\\ \Rightarrow 2\mathrm{k}+2=0\\ \Rightarrow \mathrm{k}=-1\end{array}$