$A$ line $l$ passing through the origin is perpendicular to the lines $l_{1}: \overrightarrow{r}=(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k}$ and $l_{2}: \overrightarrow{r}=(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k}$. If the coordinates of the point in the first octant on $l_{2}$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_{1}$ are $(a, b, c)$,then $18(a+b+c)$ is equal to ........ .

The vector equation of any plane passing through the line of intersection of the planes $\vec{r} \cdot \vec{m}_1=q_1$ and $\vec{r} \cdot \vec{m}_2=q_2$ is given by $\vec{r} \cdot (\vec{m}_1+\lambda \vec{m}_2)=q_1+\lambda q_2$ for $\lambda \in R$. Find the vector equation of the plane passing through the point $2 \hat{i}-3 \hat{j}+\hat{k}$ and the line of intersection of the planes $\vec{r} \cdot (\hat{i}-2 \hat{j}+3 \hat{k})=5$ and $\vec{r} \cdot (3 \hat{i}+\hat{j}-2 \hat{k})=7$.

If two distinct points $Q$ and $R$ lie on the line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$,and $PQ = PR = \sqrt{18}$,where the point $P$ is $(1, -2, 3)$,then the area of the triangle $PQR$ is equal to

The distance of the point $(1, 1, 9)$ from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x+y+z=17$ is

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:

If a plane $P$ passes through the points $(1,0,0)$ and $(0,1,0)$ and makes an angle $\frac{\pi}{4}$ with the plane $x+y=3$, then the direction ratios of a normal to that plane $P$ are