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:

Let $L$ be the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=1$. If $L$ makes an angle $\alpha$ with the positive $x$-axis,then $\cos \alpha$ equals:

Let $S$ be the reflection of a point $Q$ with respect to the plane given by $\vec{r} = -(t+p) \hat{i} + \hat{j} + (1+p) \hat{k}$,where $t, p$ are real parameters and $\hat{i}, \hat{j}, \hat{k}$ are the unit vectors along the three positive coordinate axes. If the position vectors of $Q$ and $S$ are $10 \hat{i} + 15 \hat{j} + 20 \hat{k}$ and $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ respectively,then which of the following is/are $TRUE$?
$(A)$ $3(\alpha+\beta) = -101$
$(B)$ $3(\beta+\gamma) = -71$
$(C)$ $3(\gamma+\alpha) = -86$
$(D)$ $3(\alpha+\beta+\gamma) = -121$

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$,where $\alpha > 0$,is $5$ units,then the foot of the perpendicular from $P$ to the plane is:

If $L$ is the line of intersection of two planes $x+2y+2z=15$ and $x-y+z=4$ and the direction ratios of the line $L$ are $(a, b, c)$, then $\frac{a^2+b^2+c^2}{b^2}=$

The symmetric equation of the line formed by the intersection of the planes $3x + 2y + z - 5 = 0$ and $x + y - 2z - 3 = 0$ is: