Solution

Equation of line is

$\frac{x+3}{1} = \frac{y-5}{1} = \frac{z-2}{1} = \lambda$

$\therefore$ General point $R$ on line is $R(\lambda - 3, \lambda + 5, \lambda + 2)$

$P(-2, r, 1)$

$\vec{PR} = (\lambda - 1,; \lambda + 5 - r,; \lambda + 1)$

Now $\vec{PR} \cdot \vec{d} = 0$

$\Rightarrow (\lambda - 1) + (\lambda + 5 - r) + (\lambda + 1) = 0$

$\Rightarrow 3\lambda - r + 5 = 0$

$\Rightarrow \lambda = \frac{r - 5}{3}$

$\therefore R\left(\frac{r-5}{3} - 3,; \frac{r-5}{3} + 5,; \frac{r-5}{3} + 2\right)$

$= \left(\frac{r-14}{3},; \frac{r+10}{3},; \frac{r+1}{3}\right)$

Now

$PR = \sqrt{\frac{14}{3}}$

$\Rightarrow (PR)^2 = \frac{14}{3}$

$\Rightarrow \left(\frac{r-8}{3}\right)^2 + \left(\frac{10-2r}{3}\right)^2 + \left(\frac{r-2}{3}\right)^2 = \frac{14}{3}$

$\Rightarrow r^2 - 10r + 21 = 0$

$\Rightarrow r = 3, 7$

Sum of possible values of $r = 10$