Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ with $a,b,c>0$. Since $(1,1,1)$ lies on it, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. By $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ gives $\frac{1}{3}\ge \frac{1}{(abc{)}^{1/3}}$, so $abc\ge 27$. Hence $V=\frac{abc}{6}\ge \frac{9}{2}$, with equality for $a=b=c=$ 3. So $V=\frac{9}{2}$ or $V>\frac{9}{2}$.