The angle between the two lines $\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}$ and $\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}$ is

The Cartesian equation of the line which is parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$ and passes through the point $(5, 2, -4)$ is . . . . . . .

Let a triangle $PQR$ be such that $P$ and $Q$ lie on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ and are at a distance of $6$ units from $R(1, 2, 3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\triangle PQR$, then $\alpha + \beta + \gamma$ is equal to :

The two lines $L_1: \vec{r}=(\hat{i}+5 \hat{j}+5 \hat{k})+t(4 \hat{i}-4 \hat{j}+5 \hat{k})$ and $L_2: \vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+s(8 \hat{i}-3 \hat{j}+\hat{k})$ are such that

The orthocentre of the triangle formed by the points $(2,1,5)$,$(3,2,3)$,and $(4,0,4)$ is

Let $A$ and $B$ be two distinct points on the line $L : \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of the perpendicular drawn from the point $P(1, 2, 3)$ to the line $L$. If $O$ is the origin,then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to: