The length and foot of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ are

The equation of the plane that contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ and is perpendicular to the plane $5x+3y-6z+8=0$ is

If the distance between the plane $Ax - 2y + z = d$ 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 $|d|$ is

Let $N$ be the foot of the perpendicular from the point $P(1, -2, 3)$ on the line passing through the points $(4, 5, 8)$ and $(1, -7, 5)$. Then the distance of $N$ from the plane $2x - 2y + z + 5 = 0$ is $.......$.

Let $\theta$ be the angle between the planes $P_1=\vec{r} \cdot(\hat{i}+\hat{j}+2\hat{k})=9$ and $P_2=\vec{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=15$. Let $L$ be the line that meets $P_2$ at the point $(4,-2,5)$ and makes an angle $\theta$ with the normal of $P_2$. If $\alpha$ is the angle between $L$ and $P_2$,then $(\tan^2 \theta)(\cot^2 \alpha)$ is equal to $...........$.

Find the distance of the point $(-1, -2, -1)$ from the plane passing through the point $(1, 1, 1)$ and perpendicular to both lines $L_1: \frac{x-1}{1} = \frac{y-1}{0} = \frac{z-1}{-1}$ and $L_2: \frac{x-1}{0} = \frac{y-1}{1} = \frac{z-1}{-1}$.