The foot of the perpendicular drawn from the point $(-2,-1,3)$ to a plane $\pi$ is $(1,0,-2)$. If $a, b, c$ are the intercepts made by the plane $\pi$ on $X, Y, Z$-axes respectively, then $3a+b+5c=$

The position vectors of two points $P$ and $Q$ are $3i + j + 2k$ and $i - 2j - 4k$ respectively. The equation of the plane passing through $Q$ and perpendicular to $PQ$ is:

$A$ vector $\vec{n}$ is inclined to the $x$-axis at $45^\circ$,to the $y$-axis at $60^\circ$,and at an acute angle to the $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2}, -1, 1)$,then the equation of the plane is:

The equation of the plane passing through the points $(2, 3, 1)$ and $(4, -5, 3)$ and parallel to the $X$-axis is:

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

If the foot of the perpendicular from $(0,0,0)$ to a plane is $(1,2,3)$, then the equation of the plane is