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

$A$ variable plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$,which is at a unit distance from the origin,cuts the coordinate axes at $A, B$,and $C$. If the centroid $(x, y, z)$ of $\triangle ABC$ satisfies $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=k$,then $k$ equals:

The plane $\ell x+my=0$ is rotated about its line of intersection with the plane $z=0$ through an angle $\alpha$. The equation of the new plane is

Find the distance of the point $(0, 0, 0)$ from the plane $3x - 4y + 12z = 3$. (in $/13$)

Find the equation of the plane passing through the point $(1, 2, 3)$ and parallel to the plane $2x + 3y - 4z = 0$.

If $S$ is the set of all real values of $a$ such that a plane passing through the points $(-a^2, 1, 1), (1, -a^2, 1), (1, 1, -a^2)$ also passes through the point $(-1, -1, 1)$,then $S=$