Let the line L pass through the point (-3, 5, 2) and make equal a… — Solution | JEE Mathematics

MCQMediumJEE 2026Equation of Line in 3D

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

A
$16$
B
$12$
C
$10$
D
$6$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The line $L$ passes through $A(-3,5,2)$ and makes equal angles with the positive coordinate axes. The point is $P(-2,r,1)$ and the distance from $P$ to the line is $\sqrt{\frac{14}{3}}$ .

Find: The sum of all possible values of $r$ .

A line making equal angles with the axes has direction ratios $1:1:1$ , so we take direction vector $\vec d=(1,1,1)$ .

The vector from $A$ to $P$ is

\vec{AP}=(-2-(-3),\,r-5,\,1-2)=(1,\,r-5,\,-1)

Using the distance formula from a point to a line in three dimensions,

\text{dist}^2=|\vec{AP}|^2-\frac{(\vec{AP}\cdot\vec d)^2}{|\vec d|^2}

Now,

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

and

\vec{AP}\cdot\vec d=1+(r-5)-1=r-5, \qquad |\vec d|^2=1^2+1^2+1^2=3

Therefore,

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

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

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

(r-5)^2=4

So,

r-5=\pm 2

Hence,

r=7 \text{ or } r=3

Their sum is

7+3=10

Therefore, the correct option is C.

Common mistakes

Taking the direction ratios as something other than $1:1:1$ . This is wrong because equal angles with the three positive coordinate axes imply equal direction cosines, hence proportional direction ratios $1,1,1$ . Always start by converting the geometric statement into the correct direction vector.
Using the distance formula from a point to a plane instead of from a point to a line. This gives an incorrect equation in $r$ . Here the correct relation is $\text{dist}^2=|\vec{AP}|^2-\frac{(\vec{AP}\cdot\vec d)^2}{|\vec d|^2}$ for a line with direction vector $\vec d$ .
Computing $\vec{AP}\cdot\vec d$ incorrectly. A common error is to miss the contribution of the third coordinate or mishandle signs in $1+(r-5)-1$ . Write the vector carefully as $\vec{AP}=(1,r-5,-1)$ before taking the dot product.

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