Let the line 𝐿 pass through the point (−3,5,2) and make equal angle with the positive coordinate axes. If the distance of L from the point (−2,𝑟,1) is √14/√3 , then the sum of all possible values of 𝑟 is

The correct option is 10.

Step 1: Determine the direction ratios of line L

A line that makes equal angles with the positive coordinate axes has direction cosines satisfying:

$l=m=n$

Since $l^2+m^2+n^2=1$, we have $3l^2=1$, so $l=m=n=\frac{1}{\sqrt{3}}$.

Therefore, the direction vector of line L can be taken as:

$\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$

The unit vector along line L is:

$\hat{u}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$

Step 2: Vector formulation for perpendicular distance

Let point A on line L be $A(-3,5,2)$ and point P be $P(-2,r,1)$.

The vector from A to P is given by:

$\overrightarrow{AP}=(-2-(-3\left)\right)\hat{i}+(r-5)\hat{j}+(1-2)\hat{k}=\hat{i}+(r-5)\hat{j}-\hat{k}$

The square of the distance AP is:

$|\overrightarrow{AP}{|}^2=1^2+(r-5{)}^2+(-1{)}^2=(r-5{)}^2+2$

Step 3: Calculate the projection of AP on line L

The projection length of $\overrightarrow{AP}$ along line L is:

$d=\overrightarrow{AP}·\hat{u}=\frac{1·1+(r-5)·1+(-1)·1}{\sqrt{3}}=\frac{r-5}{\sqrt{3}}$

Step 4: Use right-triangle relation for distance

The perpendicular distance $D$ from point P to line L satisfies:

$D^2=|\overrightarrow{AP}{|}^2-d^2$

We are given that $D=\frac{\sqrt{14}}{\sqrt{3}}$, so $D^2=\frac{14}{3}$.

Substituting the values into the equation:

$\frac{14}{3}=(r-5{)}^2+2-\frac{(r-5{)}^2}{3}$

Step 5: Solve for r

Combine the terms involving $(r-5{)}^2$:

$\frac{14}{3}-2=\frac{2}{3}(r-5{)}^2$

$\frac{8}{3}=\frac{2}{3}(r-5{)}^2$

Multiply both sides by $\frac{3}{2}$:

$(r-5{)}^2=4$

Taking the square root on both sides:

$r-5=2\Rightarrow r=7$

$r-5=-2\Rightarrow r=3$

Step 6: Find the sum of all possible values of r

The possible values of r are 7 and 3.

Sum = $7+3=10$.