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

If the three planes $x = 5$,$2x - 5ay + 3z - 2 = 0$,and $3bx + y - 3z = 0$ pass through a common line,then the value of $(a, b)$ is:

The point at which the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersects the plane $2x + y + z = 7$ is

The equation of the plane,passing through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to $Y$-axis is

Let $\vec{r} \cdot (3\hat{i} - 2\hat{j} + 7\hat{k}) = 32$ be the equation of a plane. If the line with direction ratios $(5, b, 3)$ is parallel to the plane, then the value of $b$ is:

If the lines $\overrightarrow{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{j} - \hat{k})$ and $\overrightarrow{r} = (\alpha\hat{i} - \hat{j}) + \mu(2\hat{i} - 3\hat{k})$ are coplanar,then the distance of the plane containing these two lines from the point $(\alpha, 0, 0)$ is