The angle between the lines $3 x=2 y=-\mathrm{z}$ and $-x=6 y=-4 z$ is

If the lines $\frac{x-1}{2}=\frac{y+1}{\mathrm{k}}=\frac{\mathrm{z}}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{\mathrm{z}}{\mathrm{k}}$ are coplanar, then the equation of the plane containing these lines are

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3 y-\alpha z+\beta=0$, then the value of $(\beta-\alpha)$ is equal to

The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3 y-6=0$, are