The equation of the plane which is parallel to the $xy$-plane and cuts an intercept of length $3$ from the $z$-axis is

$A$ plane $x$ passes through the point $(1, 1, 1)$. If $b, c, a$ are the direction ratios of a normal to the plane,where $a, b, c$ $(a < b < c)$ are the factors of $2001$,then the equation of the plane is

In $3$-dimensional space,the equation $x^2-8x+12=0$ represents ....

If the foot of the perpendicular drawn from the origin to a plane is $M(-1, -2, 2)$,then the vector equation of the plane is

Find the Cartesian equation of the plane passing through the point $(3, -3, 1)$ and normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$.

The values of $a$ for which the two points $(1, a, 1)$ and $(-3, 0, a)$ lie on the opposite sides of the plane $3x + 4y - 12z + 13 = 0$ satisfy: