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 $P$,$Q$,and $R$ are the feet of the perpendiculars drawn from the point $A(1, 1, 1)$ to the planes $P_1: x + 2y + 2z = 2$,$P_2: 2x - 2y + z = -8$,and to the line of intersection of $P_1$ and $P_2$ respectively,then the area of $\Delta PQR$ is:

The image of the point $(5, 2, 6)$ with respect to the plane $x + y + z = 9$ is

If the line segment joining the points $P(2, 4, 1)$ and $Q(3, 8, 1)$ is divided by the plane $3x - ky - 6z = 0$ externally in the ratio $4:5$,then $k=$

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to the plane:

If the distance of the point $(1, -2, 3)$ from the plane $x + 2y - 3z + 10 = 0$ measured parallel to the line $\frac{x-1}{3} = \frac{2-y}{m} = \frac{z+3}{1}$ is $\sqrt{\frac{7}{2}}$,then the value of $|m|$ is equal to ....... .