The shortest distance between the lines $$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$$ and $$\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$$ is :

If the equation of the plane containing the line

$$x+2 y+3 z-4=0=2 x+y-z+5$$ and perpendicular to the plane

$\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$

is $a x+b y+c z=4$, then $$(a-b+c)$$ is equal to :

A plane P contains the line of intersection of the plane $$\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$$ and $$\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5$$. If $$\mathrm{P}$$ passes through the point $$(0,2,-2)$$, then the square of distance of the point $$(12,12,18)$$ from the plane $$\mathrm{P}$$ is :

Let the line $$\mathrm{L}$$ pass through the point $$(0,1,2)$$, intersect the line $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$ and be parallel to the plane $$2 x+y-3 z=4$$. Then the distance of the point $$\mathrm{P}(1,-9,2)$$ from the line $$\mathrm{L}$$ is :