Let $A (1, 3, 5)$ and $B (-2, 3, -4)$ be two points. If a point $P(x, y, z)$ moves such that $PA^2 - PB^2 = 6c$,find the locus of $P$.

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 $a^2 + b^2 + c^2 =$

An equation of a plane parallel to the plane $x-2y+2z-5=0$ and which is at a distance of $1$ unit from the origin is:

$A$ plane passing through the point $(3,1,1)$ contains two lines whose direction ratios are $1, -2, 2$ and $2, 3, -1$ respectively. If this plane also passes through the point $(\alpha, -3, 5)$,then $\alpha$ is equal to

$P_1$ and $P_2$ are two distinct and intersecting planes. Three non-collinear points lie on $P_1$ and another three non-collinear points lie on $P_2$ (none being on the line of intersection of the planes). Then the maximum number of tetrahedrons formed using these six points is:

$A$ plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is