JEE Mathematics 2025 Question with Solution

Given: The point is $P(1,0,3)$ and the line passes through $A(4,7,1)$ and $B(3,5,3)$.

Find: If the image of $P$ in the given line is $Q(\alpha,\beta,\gamma)$, find $\alpha+\beta+\gamma$.

The image of a point in a line is obtained by reflecting the point across the line. So first find the foot of the perpendicular from $P$ to the line, and then use the midpoint relation.

The direction vector of the line is

Hence the vector equation of the line through $A$ is

So any point $M$ on the line is

Now,

that is,

Since $PM$ is perpendicular to the line, we use

Therefore,

Substituting $\lambda=\frac{7}{3}$ in the coordinates of $M$,

Thus, the foot of the perpendicular is

Since $M$ is the midpoint of $PQ$,

Now solve coordinate-wise:

So,

Therefore, the correct option is B, and $\alpha+\beta+\gamma=\frac{46}{3}$.

Given: $P(1,0,3)$ and line through $A(4,7,1)$, $B(3,5,3)$.

Find: The value of $\alpha+\beta+\gamma$ where $Q(\alpha,\beta,\gamma)$ is the image of $P$ in the line.

The key idea is that reflection in a line makes the line the perpendicular bisector of segment $PQ$ in 3D, so the foot point on the line becomes the midpoint of $P$ and $Q$.

Use the line through $A$ in parametric form. Its direction vector is

Hence a general point on the line is

Call this point $M$.

For $M$ to be the foot of the perpendicular from $P(1,0,3)$, vector $\vec{PM}$ must be perpendicular to the direction vector $(-1,-2,2)$. This gives one equation in $\lambda$, which yields

Therefore,

Now reflect $P$ about midpoint $M$ using

So,

Hence,

Therefore, the required sum is $\frac{46}{3}$.