Suppose the line $\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2}$ lies on the plane $x+3y-2z+\beta=0$. Then $(\alpha+\beta)$ is equal to ... .

What is the distance between the line $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 1}{2}$ and the plane $2x + 2y - z = 6$?

If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to each of the planes $3x - 2y + z = 8$ and $x + y + z = 6$ is $lx + my + nz = 1$,then $4m + 2n - 31 =$

The vector equation of a plane in parametric form, passing through the points $A(-1, 2, 0)$ and $B(2, 2, -1)$ and parallel to the line $\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{-1}$ is:

Find the equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z + 5 = 0$ and the point $(1, 1, 1)$.

If the line,$\frac{x-3}{2}=\frac{y+2}{1}=\frac{z+4}{3}$ lies in the plane,$\ell x+m y-z=9$,then $\ell^2+m^2$ is equal to