The shortest distance, between lines $$L_1$$ and $$L_2$$, where $$L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$$ and $$L_2$$ is the line, passing through the points $$\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)$$ and perpendicular to the line $$\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$$, is

Let $$L_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in \mathbb{R}$$,

$$L_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in \mathbb{R} \text {, and } L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in \mathbb{R}$$

be three lines such that $$L_1$$ is perpendicular to $$L_2$$ and $$L_3$$ is perpendicular to both $$L_1$$ and $$L_2$$. Then, the point which lies on $$L_3$$ is

Let $$(\alpha, \beta, \gamma)$$ be the foot of perpendicular from the point $$(1,2,3)$$ on the line $$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$$. Then $$19(\alpha+\beta+\gamma)$$ is equal to :

Let $$A(2,3,5)$$ and $$C(-3,4,-2)$$ be opposite vertices of a parallelogram $$A B C D$$. If the diagonal $$\overrightarrow{\mathrm{BD}}=\hat{i}+2 \hat{j}+3 \hat{k}$$, then the area of the parallelogram is equal to :