$A$ vector parallel to the line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ 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

The line given by the equations $x-2y+4z+4=0$ and $x+y+z-8=0$ intersects the plane $x-y+2z+1=0$ at the point:

Angle between the lines of intersection of the planes $x-y=0, 2x+y+z=0$ and $2x-z=0, x+y-3z=0$ is (in $^{\circ}$)

Let $\pi_1$ be a plane passing through the point $\hat{i}+\hat{j}+\hat{k}$ and perpendicular to the vector $-\hat{j}+2\hat{k}$. Let the line $L$ passing through the points $3\hat{i}-2\hat{j}+\hat{k}$ and $-\hat{i}+3\hat{j}+\hat{k}$ be a normal to the plane $\pi_2$. If the angle between the planes $\pi_1$ and $\pi_2$ is $\theta$, then $\cos \theta =$

Let the plane $P: \vec{r} \cdot \vec{a} = d$ contain the line of intersection of two planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 6$ and $\vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) = 7$. If the plane $P$ passes through the point $(2, 3, 1/2)$,then the value of $\frac{|13\vec{a}|^2}{d^2}$ is equal to