The distance of the plane $2x - 3y + 6z + 14 = 0$ from the origin is:

$A$ point on the plane determined by the points $A(1,1,-1)$, $B(2,-1,0)$, and $C(-1,0,2)$ among the following is:

Let $P(\lambda, 2, 1)$ be a point on the plane which passes through the point $Q(4, -2, 2)$. If the plane is perpendicular to the line joining the points $A(-2, -21, 29)$ and $B(-1, -16, 33)$,then find the value of $\left(\frac{\lambda}{11}\right)^{2} - \frac{4\lambda}{11} - 4$.

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them.
$2x - y + 3z - 1 = 0$ and $2x - y + 3z + 3 = 0$

$A$ plane $\pi_1$ passing through the point $3 \hat{i}-7 \hat{j}+5 \hat{k}$ is perpendicular to the vector $\hat{i}+2 \hat{j}-2 \hat{k}$ and another plane $\pi_2$ passing through the point $2 \hat{i}+7 \hat{j}-8 \hat{k}$ is perpendicular to the vector $3 \hat{i}+2 \hat{j}+6 \hat{k}$. If $p_1$ and $p_2$ are the perpendicular distances from the origin to the planes $\pi_1$ and $\pi_2$ respectively, then $p_1-p_2=$

The equation of a plane parallel to the $x$-axis is: