JEE Mathematics 2026 Question with Solution

Given: The point is $P(3,2,1)$. Its image in the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{1}$ is $Q$. We need the distance of $Q$ from the line $\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$.

Find: The required distance and the correct option.

For reflection of a point in a line, if $H$ is the foot of the perpendicular from $P$ onto the line, then

$$Q = 2H - P$$

The distance of a point from a line is

$$\text{Distance}=\frac{|\vec{BP}\times\vec{d}|}{|\vec{d}|}$$

where $\vec{d}$ is the direction vector of the line.

For the first line,

$$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{1}$$

a point on the line is $A(1,2,1)$ and its direction vector is

$$\vec{d}_1=(1,2,1)$$

Now,

$$\vec{AP}=(3-1,\,2-2,\,1-1)=(2,0,0)$$

So the projection parameter is

$$t=\frac{\vec{AP}\cdot\vec{d}_1}{|\vec{d}_1|^2} =\frac{2}{1^2+2^2+1^2} =\frac{2}{6}=\frac{1}{3}$$

Hence the foot of the perpendicular is

$$H=A+t\vec{d}_1 =\left(1+\frac{1}{3},\,2+\frac{2}{3},\,1+\frac{1}{3}\right) =\left(\frac{4}{3},\frac{8}{3},\frac{4}{3}\right)$$

Therefore the image point is

$$Q=2H-P$$

so

$$Q=\left(\frac{8}{3}-3,\,\frac{16}{3}-2,\,\frac{8}{3}-1\right) =\left(-\frac{1}{3},\,\frac{10}{3},\,\frac{5}{3}\right)$$

For the second line,

$$\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$$

a point on it is $B(9,9,5)$ and its direction vector is

$$\vec{d}_2=(3,2,-2)$$

Now,

$$\vec{BQ}=Q-B=\left(-\frac{28}{3},-\frac{17}{3},-\frac{10}{3}\right)$$

Using the cross product,

$$\vec{BQ}\times\vec{d}_2= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k} \\ -28 & -17 & -10 \\ 3 & 2 & -2 \end{vmatrix} =14\hat{i}-86\hat{j}-5\hat{k}$$

Since $\vec{BQ}$ was scaled by $3$, the actual magnitude is

$$|\vec{BQ}\times\vec{d}_2|=\frac{\sqrt{14^2+86^2+5^2}}{3}=\frac{\sqrt{7617}}{3}$$

Also,

$$|\vec{d}_2|=\sqrt{3^2+2^2+(-2)^2}=\sqrt{17}$$

Hence the distance is

$$\text{Distance}=\frac{\sqrt{7617}}{3\sqrt{17}}\approx 7$$

Therefore, the distance is $7$ and the correct option is B.