If the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}$ and $\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}$ intersect at the point $P$,then the distance of the point $P$ from the plane $z = a$ is:

The equation of the plane passing through the points $(0, 1, 2)$ and $(-1, 0, 3)$ and perpendicular to the plane $2x + 3y + z = 5$ is

Equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and has the largest distance from the origin is

Find the distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})$ and the plane $\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$.

Find the equation of the plane perpendicular to the plane containing the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and the lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$.

If two distinct points $Q$ and $R$ lie on the line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$,and $PQ = PR = \sqrt{18}$,where the point $P$ is $(1, -2, 3)$,then the area of the triangle $PQR$ is equal to