Let two planes be $P_1 : 2x - y + z = 2$ and $P_2 : x + 2y - z = 3$. Find the equation of the angle bisector plane of $P_1$ and $P_2$ that does not contain the origin.

Let the plane $x+3y-2z+6=0$ meet the coordinate axes at the points $A, B, C$. If the orthocentre of the triangle $ABC$ is $\left(\alpha, \beta, \frac{6}{7}\right)$,then $98(\alpha+\beta)^2$ is equal to $........$.

If the distance between the plane,$23x - 10y - 2z + 48 = 0$ and the plane containing the lines $\frac{x+1}{2} = \frac{y-3}{4} = \frac{z+1}{3}$ and $\frac{x+3}{2} = \frac{y+2}{6} = \frac{z-1}{\lambda}$ $(\lambda \in R)$ is equal to $\frac{k}{\sqrt{633}}$,then $k$ is equal to

$A$ variable plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$,which is at a unit distance from the origin,cuts the coordinate axes at $A, B$,and $C$. If the centroid $(x, y, z)$ of $\triangle ABC$ satisfies $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=k$,then $k$ equals:

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

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