The plane $$\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$$ cuts the $$X$$-axis at A, Y-axis at B and Z-axis at C, then the area of $$\triangle \mathrm{ABC}=$$

If a plane meets the axes $$\mathrm{X}, \mathrm{Y}, \mathrm{Z}$$ in $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ respectively such that centroid of $$\triangle \mathrm{ABC}$$ is $$(1,2,3)$$, then the equation of the plane is

The shortest distance between lines $$\bar{r}=(2 \hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$$ and $$\bar{r}=(\hat{r}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+\hat{j}-5 \hat{k})$$ is

The direction cosines $$\ell, \mathrm{m}, \mathrm{n}$$ of the line $$\frac{\mathrm{x}+2}{2}=\frac{2 \mathrm{y}-5}{3} ; \mathrm{z}=-1$$ are