The line $\frac{x - 1}{2} = -(y + 1) = \frac{z}{3}$ and the plane $3x + 2y - z = 5$ intersect at a point. The coordinates of the point are:

If $n=2 \hat{i}-3 \hat{j}+4 \hat{k}$,$m=\hat{i}-\hat{j}$,and $l=2 \hat{i}-\hat{j}+\hat{k}$,then the Cartesian equation of the plane passing through the line of intersection of two planes $r \cdot n=1$ and $r \cdot m=-4$ and perpendicular to the plane $r \cdot l=-8$ is

The angle between the line of intersection of the two planes $r \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $r \cdot(3 \hat{i}+3 \hat{j}-5 \hat{k})=3$, and the line $r=3 \hat{i}+2 \hat{j}+\hat{k}+t(5 \hat{i}+5 \hat{j}-7 \hat{k})$ is

The acute angle between the line $\frac{x-5}{2}=\frac{y+1}{-1}=\frac{z+4}{1}$ and the plane $3x-4y-z+5=0$ 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 $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$.

Let $\pi_1$ be the plane determined by the vectors $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$,and $\pi_2$ be the plane determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{j}-\hat{k}$. Let $\vec{a}$ be a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If $|\vec{a}|=\sqrt{14}$,then $|\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})|=$