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:

If the foot of the perpendicular from the point $A(-1, 4, 3)$ on the plane $P: 2x + my + nz = 4$ is $B\left(-2, \frac{7}{2}, \frac{3}{2}\right)$,then the distance of the point $A$ from the plane $P$,measured parallel to a line with direction ratios $3, -1, -4$,is equal to.

If the lines $L_1 : \frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $L_2 : \frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other, then the equation of a plane containing the line $L_1$ and parallel to the line $L_2$ for the value of $k$ that satisfies this condition is

Let the equation of the plane $P$ containing the line $x+10=\frac{8-y}{2}=z$ be $ax+by+3z=2(a+b)$ and the distance of the plane $P$ from the point $(1,27,7)$ be $c$. Then $a^2+b^2+c^2$ is equal to $.............$.

Let $P$ be a point in the first octant,whose image $Q$ in the plane $x+y=3$ (that is,the line segment $PQ$ is perpendicular to the plane $x+y=3$ and the mid-point of $PQ$ lies in the plane $x+y=3$) lies on the $z$-axis. Let the distance of $P$ from the $x$-axis be $5$. If $R$ is the image of $P$ in the $xy$-plane,then the length of $PR$ is.

The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is