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:

If $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$ lies in the plane $ax+by+z=7$, then $a+b=$

$A$ plane is making intercepts $2, 3, 4$ on $X, Y$ and $Z$-axes respectively. Another plane is passing through the point $(-1, 6, 2)$ and is perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 4)$. Then the angle between the two planes is

In $\mathbb{R}^3$,let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_1: x+2y-z+1=0$ and $P_2: 2x-y+z-1=0$. Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P_1$. Which of the following points lie$(s)$ on $M$?
$(A) \left(0, -\frac{5}{6}, -\frac{2}{3}\right)$
$(B) \left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)$
$(C) \left(-\frac{5}{6}, 0, \frac{1}{6}\right)$
$(D) \left(-\frac{1}{3}, 0, \frac{2}{3}\right)$

The distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is:

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