JEE Main 2026 — Mathematics Vectors & 3D Geometry

2026mcqmedium

If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\dfrac{x}{1} = \dfrac{y-1}{1} = \dfrac{z-2}{2}$ is $4$ , then the sum of all possible values of $a$ is equal to :

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$11$
B
$9$
C
$6$
D
$4$

Did you get this right?

Solution

Let $P(1, 2, 7)$ be the given point and the given line be $L: \dfrac{x}{1} = \dfrac{y-1}{1} = \dfrac{z-2}{2} = \lambda$ .

Any point on the line can be written as $F(\lambda, \lambda+1, 2\lambda+2)$ .

If $F$ is the foot of the perpendicular from $P$ to the line, the direction ratios of $PF$ are $(\lambda-1, \lambda-1, 2\lambda-5)$ .

Since $PF$ is perpendicular to the line, the dot product of their direction ratios is zero:

$1(\lambda-1) + 1(\lambda-1) + 2(2\lambda-5) = 0$

$6\lambda - 12 = 0 \Rightarrow \lambda = 2$

Thus, the foot of the perpendicular is $F(2, 3, 6)$ .

Let the image of $P$ in the line be $P'(x', y', z')$ . Since $F$ is the midpoint of $P$ and $P'$ :

$\dfrac{x'+1}{2} = 2 \Rightarrow x' = 3$

$\dfrac{y'+2}{2} = 3 \Rightarrow y' = 4$

$\dfrac{z'+7}{2} = 6 \Rightarrow z' = 5$

The image point is $P'(3, 4, 5)$ .

The distance of the point $Q(a, 2, 5)$ from $P'(3, 4, 5)$ is given as $4$ .

$\sqrt{(a-3)^2 + (2-4)^2 + (5-5)^2} = 4$

$(a-3)^2 + 4 = 16$

$(a-3)^2 = 12$

$a^2 - 6a - 3 = 0$

The sum of all possible values of $a$ is the sum of the roots of this quadratic equation, which is $6$ .

Answer: $6$

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Solution

Let $P(1, 2, 7)$ be the given point and the given line be $L: \dfrac{x}{1} = \dfrac{y-1}{1} = \dfrac{z-2}{2} = \lambda$ .

Any point on the line can be written as $F(\lambda, \lambda+1, 2\lambda+2)$ .

If $F$ is the foot of the perpendicular from $P$ to the line, the direction ratios of $PF$ are $(\lambda-1, \lambda-1, 2\lambda-5)$ .

Since $PF$ is perpendicular to the line, the dot product of their direction ratios is zero:

$1(\lambda-1) + 1(\lambda-1) + 2(2\lambda-5) = 0$

$6\lambda - 12 = 0 \Rightarrow \lambda = 2$

Thus, the foot of the perpendicular is $F(2, 3, 6)$ .

Let the image of $P$ in the line be $P'(x', y', z')$ . Since $F$ is the midpoint of $P$ and $P'$ :

$\dfrac{x'+1}{2} = 2 \Rightarrow x' = 3$

$\dfrac{y'+2}{2} = 3 \Rightarrow y' = 4$

$\dfrac{z'+7}{2} = 6 \Rightarrow z' = 5$

The image point is $P'(3, 4, 5)$ .

The distance of the point $Q(a, 2, 5)$ from $P'(3, 4, 5)$ is given as $4$ .

$\sqrt{(a-3)^2 + (2-4)^2 + (5-5)^2} = 4$

$(a-3)^2 + 4 = 16$

$(a-3)^2 = 12$

$a^2 - 6a - 3 = 0$

The sum of all possible values of $a$ is the sum of the roots of this quadratic equation, which is $6$ .

Answer: $6$

JEE Main 2026 — Mathematics Vectors & 3D Geometry

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JEE Main 2026 — Mathematics Vectors & 3D Geometry

Options

Did you get this right?

Solution

Your note

JEE Main Vectors & 3D Geometry in other years

More JEE Main Vectors & 3D Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Coordinate Geometry

Probability & Statistics

Trigonometry

Keep practicing Vectors & 3D Geometry