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

The equation of the plane passing through the point $(1, 2, 1)$ and perpendicular to the planes $x + 2y + 2z - 7 = 0$ and $3x + 3y + 2z - 5 = 0$ is

The plane which bisects the line segment joining the points $A(4, -2, 3)$ and $B(2, 4, -1)$ at right angles also passes through the point:

The image of the point $(1, 2, -1)$ on the plane containing the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point $(0, 7, -7)$ is:

$A$ plane at a unit distance from the origin intersects the three axes at points $P, Q,$ and $R$. If the centroid of $\Delta PQR$ is $(x, y, z)$,and it satisfies the equation $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = k$,then find the value of $k$.

Let $(\alpha, \beta, \gamma)$ be the image of the point $P (2, 3, 5)$ in the plane $2x + y - 3z = 6$. Then $\alpha + \beta + \gamma$ is equal to