JEE Mathematics 2025 Question with Solution

Given: $Q=(7,-2,5)$ and $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$. Also, $R=(5,p,q)$ lies on $L$.

Find: The square of the area of $\triangle PQR$, where $P$ is the image of $Q$ in line $L$.

Write the line in parametric form:

Since $R=(5,p,q)$ lies on $L$, use $x=5$:

Hence,

So,

To find the image point $P$ of $Q$ in the line, first find the foot $F$ of the perpendicular from $Q$ to $L$. A general point on $L$ is

Then

The direction vector of $L$ is

Since $\overrightarrow{QF} \perp \mathbf{d}$,

Therefore,

Now use reflection about the line:

So,

Form the vectors:

Use the area formula:

Now,

Hence,

Therefore,

So, the square of the area of $\triangle PQR$ is $957$.

Given: $Q=(7,-2,5)$, line $L$, and $R=(5,p,q)$ on $L$.

Find: $\left(\text{Area of } \triangle PQR\right)^2$.

A point on $L$ is $A=(1,-1,0)$ and the direction vector is $\mathbf{d}=(2,3,4)$.

From $R=(5,p,q)$ on $L$:

Thus,

Let the foot of perpendicular from $Q$ to $L$ be $F=(1+2s,-1+3s,4s)$. Then

Using perpendicularity with $\mathbf{d}=(2,3,4)$:

Hence,

Since $P$ is the reflection of $Q$ in the line, $F$ is the midpoint of $PQ$. Therefore,

Now,

Then,

So,

Therefore,

Thus, the required numerical value is $957$.