Let the plane $P$ contain the line $2x+y-z-3=0=5x-3y+4z+9$ and be parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point $A(8,-1,-19)$ from the plane $P$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to $............$.

If the lines $\frac{x - a + d}{\alpha - \delta} = \frac{y - a}{\alpha} = \frac{z - a - d}{\alpha + \delta}$ and $\frac{x - b + c}{\beta - \gamma} = \frac{y - b}{\beta} = \frac{z - b - c}{\beta + \gamma}$ are coplanar,then the equation of the plane containing them is .........

Let the lines $L_1: \vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$, $\lambda \in R$ and $L_2: \vec{r}=(4\hat{i}+\hat{j})+\mu(5\hat{i}+2\hat{j}+\hat{k})$, $\mu \in R$, intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|\overrightarrow{PR}|=\sqrt{29}$ and $|\overrightarrow{PQ}|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to

The equation of a plane passing through the intersection of two planes $x+2y-3z+2=0$ and $6x+y+z+1=0$ and parallel to the line $x-1=y+2=7-z$ is

Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+z=3$. The feet of the perpendiculars lie on the line:

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if