If the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$ cuts the co - ordinate axes at points $A, B, C$ respectively, then area of the triangle ABC is

The plane given by $\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1$ intercepts the coordinate axes at $A = (2, 0, 0)$, $B = (0, 3, 0)$, and $C = (0, 0, 6)$, forming triangle $ABC$.

Vectors representing two sides are $\vec{AB} = (-2, 3, 0)$ and $\vec{AC} = (-2, 0, 6)$.
The cross product yields $\vec{AB} \times \vec{AC} = (18, 12, 6)$ with magnitude $|\vec{AB} \times \vec{AC}| = \sqrt{18^2 + 12^2 + 6^2} = \sqrt{504} = 6\sqrt{14}$.

The area of triangle $ABC$ is $\frac{1}{2} \cdot 6\sqrt{14} = 3\sqrt{14}$ square units.
Alternatively, using the intercept formula $\frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2}$ with $a=2$, $b=3$, $c=6$ gives the same result:

$\frac{1}{2} \sqrt{(6)^2 + (18)^2 + (12)^2} = \frac{1}{2} \sqrt{504} = 3\sqrt{14}$.

Final answer: $3\sqrt{14}$ square units