The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{1}=1$ meet the co-ordinate axis at the points A, B, C respectively then the area bounded by the triangle ABC is

In the problem $A\left(2,0,0\right),B\left(0,3,0\right),C\left(0,0,1\right)$  

Area of $\Delta ABC=\frac{1}{2}\sqrt{{\Delta }_x{ }^2+{\Delta }_y{ }^2+{\Delta }_z{ }^2}$  where ${\Delta }_x=\left|\begin{array}{ccc}0& 3& 0\\ 0& 0& 1\\ 1& 1& 1\end{array}\right|=3$

${\Delta }_y=\left|\begin{array}{ccc}2& 0& 0\\ 0& 0& 1\\ 1& 1& 1\end{array}\right|=2$

${\Delta }_z=\left|\begin{array}{ccc}0& 3& 3\\ 2& 0& 0\\ 1& 1& 1\end{array}\right|=\left|(-6)\right|=6$

Required area = $\frac{1}{2}\sqrt{9+4+36}=\frac{7}{2}$