Question

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

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

(-3,5,2)

$$\overrightarrow{PR} \equiv (\lambda - 1,\lambda + 5 - r,\lambda + 1) $$Now $\overrightarrow{PR} \cdot \overrightarrow{d} = 0$

$${\Rightarrow (\lambda - 1)1 + (\lambda + 5 - r)1 + (\lambda + 1)1 = 0 }{\Rightarrow 3\lambda - r + 5 = 0 }{\Rightarrow \lambda = \frac{r - 5}{3} }{\therefore R \equiv \left( \frac{r - 5}{3} - 3,\frac{r - 5}{3} + 5,\frac{r - 5}{3} + 2 \right) }{R \equiv \left( \frac{r - 14}{3},\frac{r + 10}{3},\frac{r + 1}{3} \right) }$$Now

$${PR = \sqrt{\frac{14}{3}} \Rightarrow (PR)^{2} = \frac{14}{3} }{\Rightarrow \left( \frac{r - 14}{3} + 2 \right)^{2} + \left( \frac{r + 10}{3} - r \right)^{2} + \left( \frac{r + 1}{3} - 1 \right)^{2} = \frac{14}{3} }{\Rightarrow \frac{(r - 8)^{2}}{9} + \frac{(10 - 2r)^{2}}{9} + \frac{(r - 2)^{2}}{9} = \frac{14}{3} }{\Rightarrow \left( r^{2} - 16r + 64 \right) + \left( 100 + 4r^{2} - 40r \right) + \left( r^{2} - 4r + 4 \right) = 42 }{\Rightarrow 6r^{2} - 60r + 126 = 0 }{\Rightarrow r^{2} - 10r + 21 = 0 }{\Rightarrow r = 3,7 }$$sum of possible value of r is $= 10$