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

Find the vector equation of the plane passing through the points $R(2, 5, -3)$,$S(-2, -3, 5)$,and $T(5, 3, -3)$.

The equation of the plane passing through $(-2, 2, 2)$ and $(2, -2, -2)$ and perpendicular to the plane $9x - 13y - 3z = 0$ is

If the foot of the perpendicular drawn from the point $P(2,0,-3)$ to the plane $\pi$ is $F(1,-2,0)$ and the equation of the plane $\pi$ is $ax+by-3z+d=0$, then $a+b+d=$

If the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$ cuts the coordinate axes at points $A, B, C$ respectively,then the area of the triangle $ABC$ is

In a three-dimensional $xyz$ space,the equation $x^2 - 5x + 6 = 0$ represents: