The equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $3x+4y+5z=2$ and perpendicular to the $XY$-plane is

$A$ line drawn from the point $A(1, 3, 2)$ parallel to the line $\frac{x}{2} = \frac{y}{4} = \frac{z}{1}$ intersects the plane $3x + y + 2z = 5$ at point $B$. Then the coordinates of point $B$ are:

The point of intersection of the line passing through the points $\hat{i}-\hat{j}$ and $\hat{j}-\hat{k}$ and the plane passing through the points $2 \hat{i}+\hat{j}$, $2 \hat{j}-\hat{k}$, and $\hat{i}+2 \hat{k}$ is

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

Let $P$ be the plane containing the straight line $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$. If $d$ is the distance of $P$ from the point $(2,-5,11)$,then $d^{2}$ is equal to.

The plane $XOZ$ divides the join of $(1, -1, 5)$ and $(2, 3, 4)$ in the ratio $\lambda :1$. Then,the value of $\lambda$ is: