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 ________

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Aicharya · Accepted Answer

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

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

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…

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