The point of intersection of the line $x+1=\frac{y+3}{3}=\frac{-z+2}{2}$ with the plane $3x+4y+5z=10$ is

If $\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{k}+\hat{i}, \hat{i}-\hat{j}, \hat{j}-\hat{k}$ are the position vectors of the points $A, B, C, D, E$ respectively, then the point of intersection of the line $AB$ and the plane passing through $C, D, E$ is

$A$ straight line joining the points $(1, 1, 1)$ and $(0, 0, 0)$ intersects the plane $2x + 2y + z = 10$ at

The equation of a plane at a distance of $\sqrt{\frac{2}{21}}$ from the origin,which contains the line of intersection of the planes $x-y-z-1=0$ and $2x+y-3z+4=0$,is:

The plane $2x - y + z = 4$ intersects the line segment joining the points $A(a, -2, 4)$ and $B(2, b, -3)$ at the point $C$ in the ratio $2:1$. The distance of the point $C$ from the origin is $\sqrt{5}$. If $ab < 0$ and $P$ is the point $(a - b, b, 2b - a)$,then $CP^2$ is equal to:

If the line of intersection of the planes $ax + by = 3$ and $ax + by + cz = 0$ $(a > 0)$ makes an angle $30^{\circ}$ with the plane $y - z + 2 = 0$,then the direction cosines of the line are: