JEE Main 2025 Jan 24 Shift 2, Mathematics Q22: Lines in Space

Step 1 — Find $R$. A general point on $L$ is $(2\lambda +1,3\lambda -1,4\lambda )$. Setting the first coordinate to $5$ gives $\lambda =2$, so $R=(5,5,8)$.

Step 2 — Find the foot $T$ of the perpendicular from $Q$ to $L$. Let $T=(2\lambda +1,3\lambda -1,4\lambda )$. Then

$\overrightarrow{QT}=(2\lambda -6)\hat{i}+(3\lambda +1)\hat{j}+(4\lambda -5)\hat{k}$

Direction of $L$: $\vec{b}=2\hat{i}+3\hat{j}+4\hat{k}$. Setting $\overrightarrow{QT}\cdot \vec{b}=0$:

$2(2\lambda -6)+3(3\lambda +1)+4(4\lambda -5)=4\lambda -12+9\lambda +3+16\lambda -20=29\lambda -29=0$

$\Rightarrow \lambda =1$, so $T=(3,2,4)$.

Step 3 — Use the geometry $P$ = reflection of $Q$ in $L$.

Since $T$ is the midpoint of $QP$, $|TP|=|QT|$ and $TP\perp L$. So $QP=2QT$, and $QP\perp L$, meaning $QP\perp TR$.

Compute the two relevant lengths:

$|QT|=\sqrt{(3-7{)}^2+(2+2{)}^2+(4-5{)}^2}=\sqrt{16+16+1}=\sqrt{33}$

$|RT|=\sqrt{(3-5{)}^2+(2-5{)}^2+(4-8{)}^2}=\sqrt{4+9+16}=\sqrt{29}$

Step 4 — Area of $△PQR$.

Take $QP$ as base and the distance from $R$ to line $QP$ as height. Since $QP\perp$ direction of $L$, and both $R$ and $T$ lie on $L$, the perpendicular distance from $R$ to line $QP$ equals $|RT|=\sqrt{29}$. Base $|QP|=2|QT|=2\sqrt{33}$.

$\text{Area}=\frac{1}{2}\cdot 2\sqrt{33}\cdot \sqrt{29}=\sqrt{29\cdot 33}=\sqrt{957}$

$(\text{Area}{)}^2=957$.

Answer: 957