The distance of the plane $3x + 4y + 5z + 19 = 0$ from the point $(1, -1, 1)$ measured along a line parallel to the line with direction ratios $2, 3, 1$ is

The line $\frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 5}{4}$ lies in the plane $4x + 4y - kz - d = 0$. The values of $k$ and $d$ are

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:

The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2x-y+z+3=0$ is the line

The angle between the line $\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$ and the plane $x-2y-\lambda z=3$ is $\cos^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Then the value of $\lambda$ is:

Let the image of the point $P(1, 6, a)$ in the line $L: \frac{x-1}{2} = \frac{y-2}{b} = \frac{z-a+1}{1}, b>0$,be $Q(\frac{a}{3}, 0, a+c)$. If $S(\alpha, \beta, \gamma), \alpha > 0$,is the point on $L$ such that the distance of $S$ from the foot of perpendicular $F$ from the point $P$ on $L$ is $2\sqrt{14}$,then $\alpha + \beta + \gamma$ is equal to: