Cool Problem

Equation of the plane can be written as :

$\begin{aligned} \frac{x}{2}+\frac{y}{3} + \frac{z}{4} = 1\end{aligned}$

It cuts the $x,y,z$ axes at $(2,0,0),(0,3,0),(0,0,4)$

Let's denote the points by $\text{A(2,0,0) , B(0,3,0) , C(0,0,4)}$

By distance formula , $|AB| = \sqrt{2^2+3^2+0^2} =\sqrt{13},|BC|=\sqrt{0^2+3^2+4^2}=5,|CA|=\sqrt{2^2+0^2+4^2}=\sqrt{20}$

$cosB = \frac{AB^2+BC^2-CA^2}{2.AB.BC} = \frac{13+25-20}{10\sqrt{13}} = \frac{9}{5\sqrt{13}}$

$\Delta ABC = \frac{1}{2}|AB||BC|sinB= \frac{5\sqrt{13}}{2}\sqrt{1-\frac{81}{25.13}} = \frac{5\sqrt{13}}{2}\frac{\sqrt{244}}{5\sqrt{13}} = \frac{\cancel{5\sqrt{13}}}{\cancel{2}}\frac{\cancel{2}\sqrt{61}}{\cancel{5\sqrt{13}}} = \boxed{\sqrt{61}}$