The direction ratios of the line given by the intersection of the planes $x - y + z - 5 = 0$ and $x - 3y - 6 = 0$ are:

The line of intersection of the planes $r \cdot (i - 3j + k) = 1$ and $r \cdot (2i + 5j - 3k) = 2$ is parallel to which vector?

The equation of the line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the planes $x - 2y - z + 5 = 0$ and $x + y + 3z = 6$ is:

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. Find the equation of the plane in which they lie.

Let $\Pi$ be a plane containing the points $(0,-5,-1), (1,-2,5), (-3,5,0)$ and $L$ be a line passing through the point $(0,-5,-1)$ and parallel to the vector $\hat{i}+5\hat{j}-6\hat{k}$. Then the length of the projection of the unit normal vector to the plane $\Pi$ on the line $L$ is

In what ratio does the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 17$ divide the line segment joining the points $-2\hat{i} + 4\hat{j} + 7\hat{k}$ and $3\hat{i} - 5\hat{j} + 8\hat{k}$?