If the line $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3…
If the line $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are perpendicular to each other, then $k$ is
$\frac{7}{10}$
$\frac{10}{7}$
$\frac{-7}{10}$
$\frac{-10}{7}$
Solution
Given lines are perpendicular. Hence we write
$\begin{aligned}
&(-3)(3 \mathrm{k})+(2 \mathrm{k})(1)+(2)(-5)=0 \\
\therefore &-9 \mathrm{k}+2 \mathrm{k}-10=0 \\
\therefore & 7 \mathrm{k}=-10 \Rightarrow \quad \mathrm{k}=\frac{-10}{7}
\end{aligned}$