If the lines

The direction of ratios of the lines, $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}$, are $-3,2 k, 2$ and $3 k, 1,-5$ respectively.

It is known that two lines with direction ratios, $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$, are perpendicular, if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

$\therefore-3(3 k)+2 k \times 1+2(-5)=0$

$\Rightarrow-9 k+2 k-10=0$

$\Rightarrow 7 k=-10$

$\Rightarrow k=\frac{-10}{7}$

Therefore, for $k=-\frac{10}{7}$, the given lines are perpendicular to each other.