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

If the equation of a plane $P,$ passing through the intersection of the planes $x+4y-z+7=0$ and $3x+y+5z=8$ is $ax+by+6z=15$ for some $a, b \in R,$ then the distance of the point $(3,2,-1)$ from the plane $P$ is

Find the equation of the plane through the intersection of the planes $\vec{r} \cdot(\hat{i}+3 \hat{j})-6=0$ and $\vec{r} \cdot(3 \hat{i}-\hat{j}-4 \hat{k})=0,$ whose perpendicular distance from the origin is unity.

If $(\alpha, \beta, \gamma)$ is the intersection point of the lines $x - 3y + 2z + 4 = 0 = 2x + y + 4z + 1$ and $\frac{x - 1/3}{8} = \frac{y}{3} = \frac{z}{-1}$,then $\alpha + \beta + \gamma$ is -

If the line $\frac{x - 3}{1} = \frac{y + 2}{-1} = \frac{z + \lambda}{-2}$ lies in the plane $2x - 4y + 3z = 2$,then the shortest distance between this line and the line $\frac{x - 1}{12} = \frac{y}{9} = \frac{z}{4}$ is

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: