The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$,where $\alpha > 0$,is $5$ units,then the foot of the perpendicular from $P$ to the plane is:

Let $Q$ be the foot of the perpendicular from the point $P(7, -2, 13)$ on the plane containing the lines $\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$ and $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$. Then $(PQ)^{2}$ is equal to ..... .

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.

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 $\lambda$?

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5,$ and the point $(1,1,1).$