Let the line L pass through the point ( - 3, 5, 2) and make equal angles with the positive coordinate axes. If the distance of L from the point ( - 2, r, 1) is $$\sqrt{\frac{14}{3}}$$, then the sum of all possible values of r is:

 line $$L$$ passes through $$(-3,5,2)$$ and makes equal angles with the positive coordinate axes, so its direction vector is $$(1,1,1)$$.

The parametric equation of $$L$$ is $$(x,y,z)=(-3+t,5+t,2+t)$$.

$$A=(-3,5,2)$$ and $$P=(-2,r,1)$$ $$\vec{AP}=(-2+3,r-5,1-2)=(1,r-5,-1)$$.

The direction vector is $$\vec{d}=(1,1,1)$$ with $$|\vec{d}|=\sqrt{3}$$.

$$\vec{AP}\times\vec{d}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&r-5&-1\\1&1&1\end{vmatrix}=(r-4,-2,6-r)$$.

$$|\vec{AP}\times\vec{d}|^2=(r-4)^2+4+(6-r)^2$$.

$$\frac{(r-4)^2+4+(6-r)^2}{3}=\frac{14}{3}$$.

This gives $$(r-4)^2+(6-r)^2+4=14$$.

Expanding gives $$r^2-8r+16+r^2-12r+36=10$$.

$$2r^2-20r+52=10$$.

 $$2r^2-20r+42=0$$.
$$(r-3)(r-7)=0$$.

So $$r=3$$ or $$r=7$$.

Their sum is $$3+7=10$$.