The acute angle between the line $r = (-\hat{i} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$ and the plane $r \cdot (10\hat{i} + 2\hat{j} - 11\hat{k}) = 3$ is:

If a line $L$ is the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If line $L$ makes an angle $\alpha$ with the positive $X$-axis,then the value of $\sec \alpha$ is

The line $\frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 5}{4}$ lies in the plane $4x + 4y - kz - d = 0$. The values of $k$ and $d$ are

If for $a > 0,$ the feet of perpendiculars from the points $A(a, -2a, 3)$ and $B(0, 4, 5)$ on the plane $lx + my + nz = 0$ are points $C(0, -a, -1)$ and $D$ respectively,then the length of line segment $CD$ is equal to

$\pi$ is a plane passing through the origin and containing two lines whose direction ratios are $1, -2, 2$ and $2, 3, -1$. Then,the direction ratios of the line of intersection of the planes $x - y - z + 1 = 0$ and $\pi$ are:

The vector equation of any plane passing through the line of intersection of the planes $\vec{r} \cdot \vec{m}_1=q_1$ and $\vec{r} \cdot \vec{m}_2=q_2$ is given by $\vec{r} \cdot (\vec{m}_1+\lambda \vec{m}_2)=q_1+\lambda q_2$ for $\lambda \in R$. Find the vector equation of the plane passing through the point $2 \hat{i}-3 \hat{j}+\hat{k}$ and the line of intersection of the planes $\vec{r} \cdot (\hat{i}-2 \hat{j}+3 \hat{k})=5$ and $\vec{r} \cdot (3 \hat{i}+\hat{j}-2 \hat{k})=7$.