The line $\frac{x}{k}=\frac{y}{2}=\frac{z}{-12}$ makes an isosceles triangle with planes $2x+y+3z-1=0,$ $x+2y-3z-1=0$ then the value of k, is

The given line will be parallel to one of the bisector planes of the given planes.

$\therefore \frac{2x+y+3z-1}{\sqrt{14}}=\pm \frac{x+2y-3z-1}{\sqrt{14}}$

$3x+3y-2=0 x-y+6z=0$

$\because \frac{x}{K}=\frac{y}{2}=\frac{z}{-12}$ will be parallel to the plane $x-y+6z=0$

$\frac{1}{\frac{1}{k}}=\frac{-1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{6}{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}$

$k=-2$