Solution
\[
\begin{aligned}
& \alpha=\beta=\gamma \\
& \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \\
& \Rightarrow 3 \cos ^{2} \alpha=1 \Rightarrow \cos \alpha=\frac{1}{\sqrt{3}}
\end{aligned}
\]
direction ratios $\rightarrow 1,1,1$
\[
\begin{aligned}
& \text { line } \frac{x+3}{1}=\frac{y-5}{1}=\frac{z-2}{1}=\lambda \\
& B(\lambda-3, \lambda+5, \lambda+2) \\
& A B \xrightarrow{d r s} \lambda-3+2, \lambda+5-r, \lambda+2-1 \text { or } \lambda-1, \lambda+5-r, \lambda+1 \\
& \text { line } \xrightarrow{\mathrm{drs}} 1,1,1
\end{aligned}
\]
line $\perp \mathrm{AB} \Rightarrow(\lambda-1) 1+(\lambda+5-\mathrm{r}) 1+(\lambda+1) 1=0$
\[
\begin{aligned}
& \Rightarrow \lambda=\frac{r-5}{3} \\
& \mathrm{~B}(\lambda-3, \lambda+5, \lambda+2) \equiv\left(\frac{r-14}{3}, \frac{r+10}{3}, \frac{r+1}{3}\right) \\
& \mathrm{A}(-2, r, 1) \\
& \mathrm{AB}^{2}=\frac{14}{3} \Rightarrow\left(\frac{r-14}{3}+2\right)^{2}+\left(\frac{r+10}{3}-r\right)^{2}+\left(\frac{r+1}{3}-1\right)^{2}=\frac{14}{3} \\
& \Rightarrow\left(\frac{r-8}{3}\right)^{2}+\left(\frac{10-2 r}{3}\right)^{2}+\left(\frac{r-2}{3}\right)^{2}=\frac{14}{3} \\
& \Rightarrow r^{2}+64-16 r+100+4 r^{2}-40 r+r^{2}-4 r+4=14 \times 3=42 \\
& \Rightarrow 6 r^{2}-60 r+12 r=0 \Rightarrow r^{2}-10 r+21=0 \\
& \Rightarrow(r-7)(r-3)=0 \Rightarrow r=7,3 \text { Sum }=10 \text { Answer }(4)
\end{aligned}
\]
.
