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

Let $\lambda_1, \lambda_2$ be the values of $\lambda$ for which the points $\left(\frac{5}{2}, 1, \lambda\right)$ and $(-2, 0, 1)$ are at equal distance from the plane $2x + 3y - 6z + 7 = 0$. If $\lambda_1 > \lambda_2$,then the distance of the point $(\lambda_1 - \lambda_2, \lambda_2, \lambda_1)$ from the line $\frac{x - 5}{1} = \frac{y - 1}{2} = \frac{z + 7}{2}$ is:

If a line $L$ is common to the planes $x-y+z+2=0$ and $2x+y-2z+5=0$, then the direction cosines of the line $L$ are

$A$ ray of light passing through the point $A(1, 2, 3)$ strikes the plane $x+y+z=12$ at $B$ and on reflection passes through $C(3, 5, 9)$,then $OB$ is equal to

The point $P$ is the intersection of the straight line joining the points $Q(2, 3, 5)$ and $R(1, -1, 4)$ with the plane $5x - 4y - z = 1$. If $S$ is the foot of the perpendicular drawn from the point $T(2, 1, 4)$ to $QR$,then the length of the line segment $PS$ is

The equation of the straight line passing through $(1, 2, 3)$ and perpendicular to the plane $x + 2y - 5z + 9 = 0$ is