The equation of the plane containing the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and perpendicular to the plane containing the lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{1}$ and $\frac{x}{3}=\frac{y}{2}=\frac{z}{1}$ is

If the line joining the points $A(1,0,0)$ and $B(0,0,1)$ is a normal to the plane $\pi$ which passes through the point $A$, then the angle between the planes $\pi$ and $x+y+z=6$ is

$A$ plane $\pi$ passing through the points $2 \hat{i}-3 \hat{j}$ and $3 \hat{i}+4 \hat{k}$ is parallel to the vector $2 \hat{i}+3 \hat{j}-4 \hat{k}$. If a line joining the points $\hat{i}+2 \hat{j}$ and $\hat{j}-2 \hat{k}$ intersects the plane $\pi$ at the point $a \hat{i}+b \hat{j}+c \hat{k}$, then $a+b+2c=$

Let $A(2,5,7)$ be the image of the point $B(1,-2,3)$ with respect to a plane $\pi$. Let $C$ be the point where $AB$ meets the plane $\pi$. Let $D=(2,1,6)$. Then the direction cosines of $CD$ are

Find the equation of the plane passing through the point $(3, 2, 0)$ and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$.

If the straight lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s$ and $x = t/2, y = 1 + t, z = 2 - t$,with parameters $s$ and $t$ respectively,are coplanar,then $\lambda$ equals