The distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $r \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=4$ is

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to which of the following planes?

If the distance of the point $2 \hat{i} + 3 \hat{j} + \lambda \hat{k}$ from the plane $\vec{r} \cdot (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) = 13$ is $5$ units,then $\lambda =$

The plane $4x + 7y + 4z + 81 = 0$ is rotated through a right angle about its line of intersection with the plane $5x + 3y + 10z = 25$. The equation of the plane in its new position is $x - 4y + 6z = k$,where $k$ is:

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 $\frac{a^2+b^2+c^2}{b^2}=$

Let $O(\overrightarrow{0}), A(\hat{i}+2 \hat{j}+\hat{k}), B(-2 \hat{i}+3 \hat{k}), C(-2 \hat{i}+\hat{j}), D(4 \hat{k})$ be the position vectors of the points $O, A, B, C$ and $D$. If a line passing through $A$ and $B$ intersects the plane passing through $O, C$ and $D$ at the point $R$,then the position vector of $R$ is: