The value of $k$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$ is

If the equation of a line is $\frac{x + 3}{2} = \frac{y - 4}{3} = \frac{z + 5}{2}$ and the equation of a plane is $4x - 2y - z = 1$,then which of the following is true?

If $(\alpha, \beta, \gamma)$ is the intersection point of the lines $x - 3y + 2z + 4 = 0 = 2x + y + 4z + 1$ and $\frac{x - 1/3}{8} = \frac{y}{3} = \frac{z}{-1}$,then $\alpha + \beta + \gamma$ is -

The angle $\theta$ between the line $x - 1 = \frac{y - 2}{-1} = \frac{z - 3}{2}$ and the plane $\vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 10$ is

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

The image of the point $(5, 2, 6)$ with respect to the plane $x + y + z = 9$ is