Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The given plane is $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$

The coordinates of the points where the plane cuts the coordinate axes are $A(2,0,0)$, $B(0,3,0)$, and $C(0,0,6)$.

The vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are given by:
$\overrightarrow{AB}=-2\hat{i}+3\hat{j}+0\hat{k}$
$\overrightarrow{AC}=-2\hat{i}+0\hat{j}+6\hat{k}$

The area of $△ABC$ is $\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$.

$\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 3& 0\\ -2& 0& 6\end{array}\right|=18\hat{i}+12\hat{j}+6\hat{k}$

$|\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{18^2+12^2+6^2}=\sqrt{324+144+36}=\sqrt{504}=6\sqrt{14}$

Area of $△ABC=\frac{1}{2}\times 6\sqrt{14}=3\sqrt{14}$ sq. units.

Answer: $3\sqrt{14}$ sq. units