From a point $P(\lambda, \lambda, \lambda)$,perpendiculars $PQ$ and $PR$ are drawn respectively on the lines $y=x, z=1$ and $y=-x, z=-1$. If $P$ is such that $\angle QPR$ is a right angle,then the possible value$(s)$ of $\lambda$ is(are)

The point of intersection $C$ of the plane $8x+y+2z=0$ and the line joining the points $A(-3,-6,1)$ and $B(2,4,-3)$ divides the line segment $AB$ internally in the ratio $k:1$. If $a, b, c$ ($|a|, |b|, |c|$ are coprime) are the direction ratios of the perpendicular from the point $C$ on the line $\frac{1-x}{1}=\frac{y+4}{2}=\frac{z+2}{3}$,then $|a+b+c|$ is equal to $.............$.

Let the lines $L_{1}: \overrightarrow{r} = \lambda(\hat{i} + 2\hat{j} + 3\hat{k}), \lambda \in R$ and $L_{2}: \overrightarrow{r} = (\hat{i} + 3\hat{j} + \hat{k}) + \mu(\hat{i} + \hat{j} + 5\hat{k}), \mu \in R$ intersect at the point $S$. If a plane $ax + by - z + d = 0$ passes through $S$ and is parallel to both the lines $L_{1}$ and $L_{2}$,then the value of $a + b + d$ is equal to:

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 plane $3x - 4y - kz = 7$ contains the line $\frac{1 - x}{-2} = \frac{y + 1}{3} = \frac{z}{4}$,find the value of $k$.

The ratio in which the line joining $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is