If the point $(1, \alpha, \beta)$ lies on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}, z=1$,then $\alpha+\beta=$

If the line joining the points $\hat{i}+\hat{j}$ and $3 \hat{i}+\hat{j}-\hat{k}$ meets the plane that passes through the point $2 \hat{i}+4 \hat{j}$ and is parallel to the vectors $3 \hat{j}+5 \hat{k}$ and $3 \hat{i}-\hat{k}$ at point $P$,then the position vector of the point $P$ is

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

Let the line $L$ pass through the point $(0,1,2)$,intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$. Then the distance of the point $P(1,-9,2)$ from the line $L$ is

The mirror image of $P(2, 4, -1)$ in the plane $x - y + 2z - 2 = 0$ is $Q(a, b, c)$. Then the value of $a + b + c$ is:

Show that 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.