Let $$\mathrm{N}$$ be the foot of perpendicular from the point $$\mathrm{P}(1,-2,3)$$ on the line passing through the points $$(4,5,8)$$ and $$(1,-7,5)$$. Then the distance of $$N$$ from the plane $$2 x-2 y+z+5=0$$ is :

Let the equation of plane passing through the line of intersection of the planes $$x+2 y+a z=2$$ and $$x-y+z=3$$ be $$5 x-11 y+b z=6 a-1$$. For $$c \in \mathbb{Z}$$, if the distance of this plane from the point $$(a,-c, c)$$ is $$\frac{2}{\sqrt{a}}$$, then $$\frac{a+b}{c}$$ is equal to :

The distance of the point $$(-1,2,3)$$ from the plane $$\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10$$ parallel to the line of the shortest distance between the lines $$\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})$$ and $$\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$$ is :

Let the lines $$l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$$ and $$l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13$$ be coplanar. If the point $$\mathrm{P}(a, b, c)$$ on $$l_{1}$$ is nearest to the point $$\mathrm{Q}(-4,-3,2)$$, then $$|a|+|b|+|c|$$ is equal to