Let the line L pass through the point (-3, 5, | vedclass

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(D) The line $L$ passes through $A(-3, 5, 2)$ and makes equal angles with the coordinate axes, so its direction ratios are $(1, 1, 1)$.The equation of

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$10$

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(D) The line $L$ passes through $A(-3, 5, 2)$ and makes equal angles with the coordinate axes, so its direction ratios are $(1, 1, 1)$.The equation of the line is $\frac{x+3}{1} = \frac{y-5}{1} = \frac{z-2}{1} = \lambda$.Any general point $R$ on the line is $(\lambda-3, \lambda+5, \lambda+2)$.Let $P = (-2, r, 1)$. The vector $\overrightarrow{PR} = ((\lambda-3) - (-2), (\lambda+5) - r, (\lambda+2) - 1) = (\lambda-1, \lambda+5-r, \lambda+1)$.Since $PR$ is the perpendicular distance, $\overrightarrow{PR} \cdot \vec{d} = 0$, where $\vec{d} = (1, 1, 1)$.$(\lambda-1)(1) + (\lambda+5-r)(1) + (\lambda+1)(1) = 0 \Rightarrow 3\lambda - r + 5 = 0 \Rightarrow \lambda = \frac{r-5}{3}$.Substituting $\lambda$ back, $R = (\frac{r-5}{3}-3, \frac{r-5}{3}+5, \frac{r-5}{3}+2) = (\frac{r-14}{3}, \frac{r+10}{3}, \frac{r+1}{3})$.The distance $PR = \sqrt{\frac{14}{3}}$, so $PR^2 = \frac{14}{3}$.$PR^2 = (\frac{r-14}{3} + 2)^2 + (\frac{r+10}{3} - r)^2 + (\frac{r+1}{3} - 1)^2 = \frac{14}{3}$.$(\frac{r-8}{3})^2 + (\frac{10-2r}{3})^2 + (\frac{r-2}{3})^2 = \frac{14}{3}$.$\frac{r^2-16r+64 + 100-40r+4r^2 + r^2-4r+4}{9} = \frac{14}{3}$.$6r^2 - 60r + 168 = 42 \Rightarrow 6r^2 - 60r + 126 = 0$.$r^2 - 10r + 21 = 0 \Rightarrow (r-3)(r-7) = 0$.The possible values of $r$ are $3$ and $7$.The sum of all possible values of $r$ is $3 + 7 = 10$.

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 $P(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is:

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