The point of intersection of the line $x+1=\frac{y+3}{3}=\frac{-z+2}{2}$ with the plane $3x+4y+5z=10$ is

On which of the following lines does the point of intersection of the line $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ lie?

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

If ${L_1}$ is the line of intersection of the planes $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$,and ${L_2}$ is the line of intersection of the planes $x + 2y - z - 3 = 0$ and $3x - y + 2z - 1 = 0$,then the distance of the origin from the plane containing the lines ${L_1}$ and ${L_2}$ is:

Let $P_1$ be the plane $3x - y - 7z = 11$ and $P_2$ be the plane passing through the points $(2, -1, 0)$,$(2, 0, -1)$,and $(5, 1, 1)$. If the foot of the perpendicular drawn from the point $(7, 4, -1)$ on the line of intersection of the planes $P_1$ and $P_2$ is $(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma$ is equal to $............$.

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: