The line $\frac{x + 3}{3} = \frac{y - 2}{-2} = \frac{z + 1}{1}$ and the plane $4x + 5y + 3z - 5 = 0$ intersect at a point

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:

For a positive real number $p$,if the perpendicular distance from a point $-\hat{i} + p\hat{j} - 3\hat{k}$ to the plane $\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 7$ is $6$ units,then $p=$

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 4$ and $2x + y - z = -5$,and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.

The distance of the point $P(3,4,4)$ from the point of intersection of the line joining the points $Q(3,-4,-5)$ and $R(2,-3,1)$ and the plane $2x+y+z=7$ is equal to $.....$

If the line $\frac{x - 3}{1} = \frac{y + 2}{-1} = \frac{z + \lambda}{-2}$ lies in the plane $2x - 4y + 3z = 2$,then the shortest distance between this line and the line $\frac{x - 1}{12} = \frac{y}{9} = \frac{z}{4}$ is