If the foot of the perpendicular drawn from the point $P(2,0,-3)$ to the plane $\pi$ is $F(1,-2,0)$ and the equation of the plane $\pi$ is $ax+by-3z+d=0$, then $a+b+d=$

If planes $x - c y - b z = 0$,$c x - y + a z = 0$ and $b x + a y - z = 0$ pass through a straight line then $a^2 + b^2 + c^2 =$

The Cartesian equation of the plane passing through the point $(1, -2, 3)$ and perpendicular to the vector $-\hat{i} + 2\hat{j} - 3\hat{k}$ is:

If a plane meets the coordinate axes at $A, B$ and $C$ in such a way that the centroid of $\triangle ABC$ is at the point $(1, 2, 3)$,then the equation of the plane is

Let the plane $\pi$ pass through the point $(1,0,1)$ and be perpendicular to the planes $2x+3y-z=2$ and $x-y+2z=1$. Let the equation of the plane passing through the point $(11,7,5)$ and parallel to the plane $\pi$ be $ax+by-z-d=0$. Then,$\frac{a}{b}+\frac{b}{d}=$

If the equation of a plane passing through $A(1, p, 2)$ and $B(3, 2, 4)$ and parallel to the $z$-axis is $3x - 2y - q = 0$, then: