Mathematics - Vector Question with Solution | TestHub

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

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