The equation of the plane passing through a point $A(2, -1, 3)$ and parallel to the vectors $\vec{a} = (3, 0, -1)$ and $\vec{b} = (-3, 2, 2)$ is:

Assertion: The points $(2, 1, 5)$ and $(3, 4, 3)$ lie on opposite sides of the plane $2x + 2y - 2z - 1 = 0$.
Reason: The algebraic perpendicular distances from the given points to the plane have opposite signs.

Find the vector and Cartesian equations of the plane that passes through the point $(1, 4, 6)$ and the normal vector to the plane is $\hat{i} - 2\hat{j} + \hat{k}$.

The equation of the plane passing through $(4,4,0)$ and perpendicular to the planes $2x+y+2z+3=0$ and $3x+3y+2z-8=0$ is

$P_1$ 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:

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