JEE Main 2025 — Mathematics Vectors & 3D Geometry

$\begin{aligned}
& \mathrm{L}_1: \mathrm{x}+2=\mathrm{y}-1=\mathrm{z}=\ell \
& \mathrm{L}_2: \frac{\mathrm{x}-3}{5}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\mathrm{m} \
& \mathrm{~L}_3: \frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-3}{5}=\frac{\mathrm{z}-2}{1}=\mathrm{n}
\end{aligned}$

Point of intersection of $L_1$L1​ and $L_2$L2​

$\left.\begin{array}{r}
\ell-2=5 \mathrm{m}+3 \
\ell+1=-\mathrm{m} \
\ell=\mathrm{m}+1
\end{array}\right} \ell=0, \mathrm{m}=-1 \quad \mathrm{~A}(-2,1,0)$

Point of intersection of $L_2$L2​ and $L_3$L3​

$\left.\begin{array}{l}
5 \mathrm{m}+3=-3 \mathrm{n} \
-\mathrm{m}=3 \mathrm{n}+3 \
\mathrm{m}+1=\mathrm{n}+2
\end{array}\right} \mathrm{m}=0, \mathrm{n}=-1, B(3,0,1)$

Point of intersection $L_3$L3​ and $L_4$L4​

$\left.\begin{array}{r}
-3 \mathrm{n}=\ell-2 \
3 \mathrm{n}+3=\ell+1 \
\mathrm{n}+2=\ell
\end{array}\right} \ell=2, \mathrm{n}=0, \mathrm{C}(0,3,2)$

$$\begin{aligned} & \operatorname{Ar}(\Delta \mathrm{ABC})=\left|\frac{1}{2}\right| \begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -5 & 1 & -1 \\ -3 & 3 & 1\end{array}| | \\ & \mathrm{A}=\frac{1}{2}|\hat{\mathrm{i}}(4)-\hat{\mathrm{j}}(-8)+\hat{\mathrm{k}}(-12)| \\ & \mathrm{A}=\frac{1}{2} \sqrt{16+64+144}=\sqrt{56} \\ & \mathrm{~A}^2=56\end{aligned}$$​Ar(ΔABC)=​21​​i^−5−3​j^​13​k^−11​∣∣A=21​∣i^(4)−j^​(−8)+k^(−12)∣A=21​16+64+144​=56​ A2=56​