In space,the equation $by + cz + d = 0$ represents a plane perpendicular to the

Let the plane passing through the point $(2,1,-1)$ and containing the line joining the points $(1,3,2)$ and $(1,2,1)$ make intercepts $p, q, r$ on the coordinate axes. Then $p+q+r=$

Two systems of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a', b', c'$ from the origin along the axes,then:

If the mirror image of the point $(2, 4, 7)$ in the plane $3x - y + 4z = 2$ is $(a, b, c)$,then $2a + b + 2c$ is equal to

$A$ plane $\pi$ passing through the point $(1,1,1)$ is perpendicular to the line joining the points $(6,3,2)$ and $(1,-4,-9)$. If $ax+by+cz-23=0$ is the equation of the plane $\pi$,then $a+b-c=$

$A$ plane passes through $(1, -2, 1)$ and is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. The distance of the point $(1, 2, 2)$ from this plane is . . . . . . units.