Let in a $\triangle ABC$,the length of the side $AC$ be $6$,the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is

The shortest distance between the skew lines $\vec{r}=(2 \hat{i}-\hat{j})+t(\hat{i}+2 \hat{k})$ and $\vec{r}=(-2 \hat{i}+\hat{k})+s(\hat{i}-\hat{j}-\hat{k})$ is

The shortest distance between the skew lines $\vec{r}=(-\hat{i}-2 \hat{j}-3 \hat{k})+t(3 \hat{i}-2 \hat{j}-2 \hat{k})$ and $\vec{r}=(7 \hat{i}+4 \hat{k})+s(\hat{i}-2 \hat{j}+2 \hat{k})$ is

Let $A(2,3,5), B(-1,3,2), C(\lambda, 5, \mu)$ be the vertices of $\triangle ABC$. If the median through the vertex $A$ is equally inclined to the coordinate axes,then

$L_1$ is a line passing through the points with position vectors $\hat{i}-2 \hat{j}-\hat{k}$ and $4 \hat{i}-3 \hat{k}$. $L_2$ is a line passing through the points with position vectors $\hat{i}+2 \hat{j}-\hat{k}$ and $2 \hat{i}-4 \hat{j}-5 \hat{k}$. Then the distance between $L_1$ and $L_2$ is

If the image of the point $(4,4,3)$ in the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to