The position vector of the point of intersection of the line $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and the $XOY$-plane is:

The equation of the plane,passing through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to $Y$-axis is

Let $L$ be a line passing through the points $2 \hat{i}+3 \hat{j}+8 \hat{k}$ and $\hat{i}+6 \hat{j}+4 \hat{k}$. Let $P$ be a plane passing through $-5 \hat{i}+19 \hat{j}-14 \hat{k}$ and parallel to the vectors $\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-2 \hat{j}+3 \hat{k}$. If $L$ meets the plane $P$ at a point $A$, then the position vector of $A$ is:

If the distance between the plane $ax - 2y + z = k$ and the plane containing the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$, then $|k|$ is

The line of intersection of the planes $x + 2y = 0$ and $y - 3z + 3 = 0$ is

$\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=3$ are two planes. $A$ plane $\pi$ passing through the line of intersection of these two planes, passes through the point $(0,1,2)$. If the equation of $\pi$ is $\vec{r} \cdot(a \hat{i}+b \hat{j}+c \hat{k})=m$, then $\frac{b c}{a^2}=$