The equation of the plane $\pi$ through the line of intersection of the planes $\pi_1 \equiv x+3y-6=0$ and $\pi_2 \equiv 3x-y+4z=0$ is $\pi_1+\lambda \pi_2=0$. If the plane $\pi$ is at unit distance from the origin,then an equation of the plane $\pi$ is

The equation of the plane passing through the intersection of planes $2x - y + z = 3$ and $4x - 3y + 5z = -9$ and parallel to the line $\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z - 3}{5}$ is

The coordinates of the point where the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersects the plane $2x + y + z = 7$ are:

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 $\cos \theta =$

If the equation of the plane passing through the point $(1, 1, 2)$ and perpendicular to the intersection of the planes $x - 3y + 2z - 1 = 0$ and $4x - y + z = 0$ is $Ax + By + Cz = 1$,then $140(C - B + A)$ is equal to $.........$.

On which of the following lines does the point of intersection of the line $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ lie?