Let the points on the plane $P$ be equidistant from the points $A(-4, 2, 1)$ and $B(2, -2, 3)$. Then the acute angle between the plane $P$ and the plane $2x + y + 3z = 1$ is

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:

The equation of a plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by

Find the distance from the origin to the plane passing through the point $(2, 3, -1)$ and perpendicular to the vector $3\hat{i} - 4\hat{j} + 7\hat{k}$.

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

The $x$-intercept of a plane $\pi$ passing through the point $(1,1,1)$ is $\frac{5}{2}$ and the perpendicular distance from the origin to the plane $\pi$ is $\frac{5}{7}$. If the $y$-intercept of the plane $\pi$ is negative and the $z$-intercept is positive,then its $y$-intercept is