The equation of a plane containing the line $x-2=\frac{y-4}{4}=\frac{z-6}{7}$ and parallel to the line $\vec{r}=(\hat{i}+3\hat{j}+5\hat{k})+\lambda(3\hat{i}+5\hat{j}+7\hat{k})$ is

The distance of the point $(2, 3, 4)$ from the plane $3x - 6y + 2z + 11 = 0$ is

The plane through the intersection of planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to $Y$-axis also passes through the point

If $P=(2,-3,4)$,$Q=(-1,-4,0)$,and $R=(2,1,0)$ are three points,and $S$ is the foot of the perpendicular drawn from $R$ to the line $PQ$,then the $X$-coordinate of $S$ is:

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7$ and $\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$ and passing through the point $(2,1,3).$

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