The equation of the plane containing the line $\frac{x+1}{2}=\frac{y+2}{1}=\frac{z-2}{3}$ and the point $(1,-1,3)$ is

The equation of the plane in which the lines $\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{-5}$ and $\frac{x - 8}{7} = \frac{y - 4}{1} = \frac{z - 5}{3}$ lie,is

The equation of the plane passing through the line of intersection of planes $\pi_1: 2x + 6y + 4z - 7 = 0$ and $\pi_2: x - y - 2z - 2 = 0$, and perpendicular to the plane $x + y + 2z - 5 = 0$ is:

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is

Let the equation of the plane,that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$,be $\alpha x+\beta y+\gamma z+3=0$. Then $\alpha+\beta+\gamma$ is equal to:

The distance of the point $(-1, 2, 3)$ from the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 10$ measured parallel to the line of the shortest distance between the lines $\vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j}) + \mu(\hat{i} - \hat{j} + \hat{k})$ is: