Find the distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $\vec{r} \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=9$.

The vector equation of a plane passing through the line of intersection of the planes $\overline{r} \cdot(\overline{i}-2 \overline{k})=3$ and $\overline{r} \cdot(2 \overline{j}+\overline{k})=5$,and passing through the point $\overline{i}+2 \overline{j}+3 \overline{k}$,is:

If the line $\frac{x-3}{2}=\frac{y+5}{-1}=\frac{z+2}{2}$ lies in the plane $\alpha x+3y-z+\beta=0$,then the values of $\alpha$ and $\beta$ respectively are ....

Find the position vector of the point where the line $r = (i - j + k) + t(i + j + k)$ meets the plane $r \cdot (i + j - k) = 5$.

Let $P$ be the plane passing through the intersection of the planes $\overrightarrow{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\overrightarrow{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$,and the point $(2, 1, -2)$. Let the position vectors of the points $X$ and $Y$ be $\hat{i} - 2\hat{j} + 4\hat{k}$ and $5\hat{i} - \hat{j} + 2\hat{k}$ respectively. Then the points:

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^{\circ}$ about its line of intersection with the plane $x + y + z = 1$. The equation of the plane in its new position is: