Let the line $\frac{x - 2}{3} = \frac{y - 1}{-5} = \frac{z + 2}{2}$ lie in the plane $x + 3y - \alpha z + \beta = 0$. Then $(\alpha, \beta)$ equals.

If the plane $P$ passes through the intersection of two mutually perpendicular planes $2x + ky - 5z = 1$ and $3kx - ky + z = 5$,where $k < 3$,and intercepts a unit length on the positive $x$-axis,then the intercept made by the plane $P$ on the $y$-axis is

The perpendicular distance from the origin to the plane containing the two lines,$\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$,is

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 -

If the point of intersection of the lines $r = \hat{i} - 6\hat{j} + (p \sec \alpha) \hat{k} + t(\hat{i} + 2\hat{j} + \hat{k})$ and $r = 4\hat{j} + \hat{k} + \lambda(2\hat{i} + (p \tan \alpha) \hat{j} + 2\hat{k})$ is $8\hat{i} + 8\hat{j} + 9\hat{k}$, (where $0 < \alpha < \frac{\pi}{2}$), then $p =$

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 the $X$-axis is