If the point of intersection of the lines $\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7}$ and $\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7}$ lies on the $xy$-plane, then the value of $a+b$ is:

The distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$ is

The equation of the plane passing through the intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to the $x$-axis is:

The equation of the plane passing through the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ and perpendicular to the plane $x+2y+z=12$ is given by $ax+by+cz+4=0$. Then:

Find the equation of the line passing through $(1, 1, 1)$ and perpendicular to the plane $2x + 3y - z - 5 = 0$.

$A$ plane $\pi$ is passing through the points $A(1, -2, 3)$ and $B(6, 4, 5)$. If the plane $\pi$ is perpendicular to the plane $3x - y + z = 2$,then the perpendicular distance from $(0, 0, 0)$ to the plane $\pi$ is