Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ intersects the coordinate axes at $A(2,0,0)$, $B(0,3,0)$, and $C(0,0,4)$.

The area of $\Delta ABC$ is given by $\Delta =\sqrt{{\Delta }_{xy}^2+{\Delta }_{yz}^2+{\Delta }_{zx}^2}$, where ${\Delta }_{xy}$, ${\Delta }_{yz}$, and ${\Delta }_{zx}$ are the areas of the projections of $\Delta ABC$ on the $xy$, $yz$, and $zx$ planes respectively.

${\Delta }_{xy}=\frac{1}{2}\times 2\times 3=3$

${\Delta }_{yz}=\frac{1}{2}\times 3\times 4=6$

${\Delta }_{zx}=\frac{1}{2}\times 4\times 2=4$

$\Delta =\sqrt{3^2+6^2+4^2}=\sqrt{9+36+16}=\sqrt{61}$

Answer: $\sqrt{61}$