If the equation of the plane passing through the points $(2,1,2)$ and $(1,2,1)$ and perpendicular to the plane $2x - y + 2z = 1$ is $ax + by + cz + d = 0$, then $\frac{a+b}{c+d} = $

$A$ plane $E$ is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,and passes through the point $P(1, -1, 1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3\sqrt{2}$,then $(PQ)^2$ is equal to

The distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is:

If the equation of the plane passing through the point $(1, 1, 2)$ and perpendicular to the intersection of the planes $x - 3y + 2z - 1 = 0$ and $4x - y + z = 0$ is $Ax + By + Cz = 1$,then $140(C - B + A)$ is equal to $.........$.

If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to the planes $3x - 2y + z = 9$ and $x + y + z = 9$ is $x + by + cz + d = 0$, then $d =$

The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{-5}$ in the plane $2x - y + z + 3 = 0$ is the line: