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:

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

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

We seek the distance from the point $$(-2,r,1)$$ to the line $$L$$.

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

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

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

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

Thus the square of the distance is $$\frac{(r-4)^2+4+(6-r)^2}{3}=\frac{14}{3}$$.

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

Rewriting yields $$(r-4)^2+(r-6)^2=10$$.

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

Combining terms leads to $$2r^2-20r+52=10$$.

Hence $$2r^2-20r+42=0$$.

Dividing by 2 gives $$r^2-10r+21=0$$.

Factoring yields $$(r-3)(r-7)=0$$.

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

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

The answer is Option 1: 10.