Exam

JEE Main

Year

2025

Date

8 April

Shift

Evening Shift

Subject

Mathematics

Chapter

Three Dimensional Geometry

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Three Dimensional Geometry

Let the area of the triangle formed by the lines x+2=y-1=z,...

Let the area of the triangle formed by the lines $x+2=y-1=z, \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to ________

Step-by-step solution

$$\begin{aligned}<br/>& \mathrm{L}_1: \mathrm{x}+2=\mathrm{y}-1=\mathrm{z}=\ell \\ & \mathrm{L}_2: \frac{\mathrm{x}-3}{5}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\mathrm{m} \\ & \mathrm{~L}_3: \frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-3}{5}=\frac{\mathrm{z}-2}{1}=\mathrm{n}<br/>\end{aligned}$$$<br/>Point of intersection of$L_1 and L_2<br/>\left.$\begin{array}{r}<br/>\ell-2=5 \mathrm{~m}+3 \\ \ell+1=-\mathrm{m} \\ \ell=\mathrm{m}+1<br/>\end{array}$\right\} \ell=0, \mathrm{~m}=-1 \quad \mathrm{~A}(-2,1,0)
Point of intersection of $L_2 and L_3$
\left.$$\begin{array}{l}<br/>5 \mathrm{~m}+3=-3 \mathrm{n} \\ -\mathrm{m}=3 \mathrm{n}+3 \\ \mathrm{~m}+1=\mathrm{n}+2<br/>\end{array}$$\right\} \mathrm{m}=0, \mathrm{n}=-1, B(3,0,1)$<br/>Point of intersection$L_3 and L_4<br/>\left.$\begin{array}{r}<br/>-3 \mathrm{n}=\ell-2 \\ 3 \mathrm{n}+3=\ell+1 \\ \mathrm{n}+2=\ell<br/>\end{array}$\right\} \ell=2, \mathrm{n}=0, \mathrm{C}(0,3,2)

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