If the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1,$ cuts the coordinates axes in A, B, C, then the area of ΔABC is

$\sqrt{61}$ sq.units

The given place cuts the coordinate axes in A(2, 0, 0), B(0, 3, 0) and C(0, 0, 4).

∴ Area of ΔABC $=\frac{1}{2}AB×AC × sin ∠BAC$

Now, 

$AB = \sqrt{4 + 9 + 0}= \sqrt{13}, AC = \sqrt{4+0+16}= \sqrt{20}.$

$cos∠BAC \frac{\vec{AB}.\vec{AC}}{|\vec{AB}||\vec{AC}|}=\frac{(-2\hat{i}+3\hat{j}).(-2\hat{i}+4\hat{k})}{\sqrt{4+9}\sqrt{4+16}}$

$⇒cos∠BAC =\frac{4+0+0}{\sqrt{13}\sqrt{20}}=\frac{4}{\sqrt{13}\sqrt{20}}=\frac{2}{\sqrt{65}}$

$⇒sin∠BAC= \sqrt{1-\frac{4}{65}}=\sqrt{\frac{61}{65}}$

Hence,

Area of ΔABC = $\frac{1}{2}×\sqrt{13}×\sqrt{20} ×\sqrt{\frac{61}{65}}= \sqrt{61}$ sq.units.