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

If the line $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2x - 4y + z = 7$,then $k = \dots$

If for $a > 0,$ the feet of perpendiculars from the points $A(a, -2a, 3)$ and $B(0, 4, 5)$ on the plane $lx + my + nz = 0$ are points $C(0, -a, -1)$ and $D$ respectively,then the length of line segment $CD$ is equal to

The equation of the plane containing the line $\frac{x+1}{2}=\frac{y+2}{1}=\frac{z-2}{3}$ and the point $(1,-1,3)$ is

If the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z + 4}{3}$ lies in the plane $lx + my - z = 9$,then $l^2 + m^2 = \dots$

The value of $\lambda$ for which the straight line $\frac{x-\lambda}{3}=\frac{y-1}{2+\lambda}=\frac{z-3}{-1}$ may lie on the plane $x-2y=0$ is