JEE Mathematics 2026 Question with Solution

Given: The image of $P(a,2,a)$ in line $L_1: \frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}$ is $Q$, and the image of $Q$ in line $L_2: \frac{x-2b}{2}=\frac{y-a}{1}=\frac{z+2b}{-5}$ is $P$.

Find: The value of $a+b$.

If one point is the image of the other in a line, that line is the perpendicular bisector of the segment joining the two points. Therefore, both lines are perpendicular bisectors of $PQ$, so they must intersect at the midpoint $M$ of $PQ$.

A general point on $L_1$ is

$$M_1=(2\lambda,\lambda-a,\lambda)$$

and a general point on $L_2$ is

$$M_2=(2\mu+2b,\mu+a,-5\mu-2b).$$

Since these represent the same intersection point, equate coordinates:

$$2\lambda=2\mu+2b \implies \lambda=\mu+b$$ $$\lambda-a=\mu+a \implies \lambda=\mu+2a$$

Comparing these gives

$$b=2a.$$

Now equate the $z$-coordinates:

$$\lambda=-5\mu-2b.$$

Using $\lambda=\mu+2a$ and $b=2a$,

$$\mu+2a=-5\mu-4a$$ $$6\mu=-6a \implies \mu=-a.$$

Hence,

$$\lambda=\mu+2a=a.$$

So the intersection point is

$$M=(2a,0,a).$$

Since $M$ is the midpoint and $L_1$ is the perpendicular bisector, vector $\overrightarrow{MP}$ is perpendicular to the direction vector of $L_1$.

Now

$$P=(a,2,a), \quad M=(2a,0,a)$$

so

$$\overrightarrow{MP}=P-M=(-a,2,0).$$

The direction vector of $L_1$ is

$$\vec d_1=(2,1,1).$$

Using perpendicularity,

$$\overrightarrow{MP}\cdot \vec d_1=0$$ $$(-a,2,0)\cdot(2,1,1)=0$$ $$-2a+2=0 \implies a=1.$$

Therefore,

$$b=2a=2.$$

Hence,

$$a+b=1+2=3.$$

Therefore, the required value is $3$. The solution's lists $14$, but the extracted working clearly gives $3$.

Given: Both reflections connect the same pair of points $P$ and $Q$.

Find: $a+b$.

The key idea is that if $Q$ is the reflection of $P$ in a line, then that line is the perpendicular bisector of $PQ$. Since both given lines perform reflection between the same two points, both must pass through the midpoint of $PQ$.

So first find the intersection point of the two lines.

From

$$\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}=\lambda,$$

we get

$$x=2\lambda, \quad y=\lambda-a, \quad z=\lambda.$$

From

$$\frac{x-2b}{2}=\frac{y-a}{1}=\frac{z+2b}{-5}=\mu,$$

we get

$$x=2\mu+2b, \quad y=\mu+a, \quad z=-5\mu-2b.$$

At the intersection,

$$2\lambda=2\mu+2b,$$ $$\lambda-a=\mu+a,$$ $$\lambda=-5\mu-2b.$$

From the first two equations,

$$\lambda=\mu+b,$$ $$\lambda=\mu+2a.$$

Hence,

$$b=2a.$$

Substitute this into the third equation:

$$\mu+2a=-5\mu-4a$$ $$6\mu=-6a$$ $$\mu=-a.$$

Then

$$\lambda=\mu+2a=a.$$

So the midpoint is

$$M=(2a,0,a).$$

Now use the fact that $PM$ must be perpendicular to $L_1$.

We have

$$\overrightarrow{MP}=(-a,2,0)$$

and direction vector of $L_1$ is

$$(2,1,1).$$

Thus,

$$(-a,2,0)\cdot(2,1,1)=0$$ $$-2a+2=0$$ $$a=1.$$

Then

$$b=2.$$

Therefore,

$$a+b=3.$$

So the correct numerical answer from the shown working is $3$.