Find the equation of the plane passing through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2x + 6y + 6z - 1 = 0$.

The direction ratios of the two lines $AB$ and $AC$ are $1, -1, -1$ and $2, -1, 1$. The direction ratios of the normal to the plane $ABC$ are

The plane passing through the point $(4, -1, 2)$ and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{-1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point

The equation of a plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

Statement $-1:$ The point $A(3,1,6)$ is the mirror image of the point $B(1,3,4)$ in the plane $x-y+z=5.$
Statement $-2:$ The plane $x-y+z=5$ bisects the line segment joining $A(3,1,6)$ and $B(1,3,4).$

Let the mirror image of the point $P(1, 3, a)$ with respect to the plane $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) - b = 0$ be $Q(-3, 5, 2)$. Then the value of $|a + b|$ is equal to ......