The equation of the plane that contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ and is perpendicular to the plane $5x+3y-6z+8=0$ is

Find the equation of the plane passing through the points $(2, 1, -1)$ and $(-1, 3, 4)$ and perpendicular to the plane $x - 2y + 4z = 10$.

If $L$ is the line of intersection of two planes $x+2y+2z=15$ and $x-y+z=4$ and the direction ratios of the line $L$ are $(a, b, c)$, then $\frac{a^2+b^2+c^2}{b^2}=$

The lines $\frac{x-0}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda}$ and $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{\lambda}$ are coplanar. If $p$ is the plane containing these lines,then which of the following points lies on the plane for all values of $\lambda$?

If the line joining the points $\overline{i} + 2\overline{j}$ and $\overline{j} - 2\overline{k}$ intersects the plane passing through the points $2\overline{i} - \overline{j}$,$2\overline{j} + 3\overline{k}$,and $\overline{k} - 2\overline{i}$ at $\overline{r}$,then $\overline{r} \cdot (\overline{i} + \overline{j} + \overline{k}) = $

$A$ plane $P$ meets the coordinate axes at $A, B$ and $C$ respectively. The centroid of $\Delta ABC$ is given to be $(1, 1, 2)$. Then the equation of the line through this centroid and perpendicular to the plane $P$ is