The position vector of the point of intersection of the line $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and the $XOY$-plane is:

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 line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4}$ meets the plane $x + 2y + 3z = 14$ at the point:

The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is:

Let the equation of the plane passing through the line of intersection of the planes $x+2y+az=2$ and $x-y+z=3$ be $5x-11y+bz=6a-1$. For $c \in \mathbb{Z}$,if the distance of this plane from the point $(a, -c, c)$ is $\frac{2}{\sqrt{a}}$,then $\frac{a+b}{c}$ is equal to

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{b} = 2\hat{i} + 7\hat{j} + 3\hat{k}$. Let $L_1: \overrightarrow{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \overrightarrow{a}, \lambda \in R$ and $L_2: \overrightarrow{r} = (\hat{j} + \hat{k}) + \mu \overrightarrow{b}, \mu \in R$ be two lines. If the line $L_3$ passes through the point of intersection of $L_1$ and $L_2$,and is parallel to $\overrightarrow{a} + \overrightarrow{b}$,then $L_3$ passes through the point: