Answer
$\frac{9}{2}$ is the minimum value of $V$ among all planes passing through the point $P = \left( {1,1,1} \right)$.
Work Step by Step
We are given a tetrahedron formed by the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ ($a,b,c > 0$) with the positive coordinate planes $x,y,z > 0$.
Our task is to find $a$, $b$, and $c$ such that the volume of the tetrahedron $V = \frac{1}{6}abc$ is minimum among all planes passing through the point $P = \left( {1,1,1} \right)$.
Since the plane must pass through $P = \left( {1,1,1} \right)$, the equation of the plane becomes
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$
Thus, our problem can be stated: minimize $V = \frac{1}{6}abc$ subject to the constraint $g\left( {a,b,c} \right) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla V = \lambda \nabla g$ yields
$\left( {\frac{{bc}}{6},\frac{{ac}}{6},\frac{{ab}}{6}} \right) = \lambda \left( { - \frac{1}{{{a^2}}}, - \frac{1}{{{b^2}}}, - \frac{1}{{{c^2}}}} \right)$
So, the Lagrange equations are
(1) ${\ \ \ \ }$ $\frac{{bc}}{6} = - \frac{\lambda }{{{a^2}}}$, ${\ \ }$ $\frac{{ac}}{6} = - \frac{\lambda }{{{b^2}}}$, ${\ \ }$ $\frac{{ab}}{6} = - \frac{\lambda }{{{c^2}}}$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $a,b,c > 0$, so
$\lambda = - \frac{{{a^2}bc}}{6} = - \frac{{a{b^2}c}}{6} = - \frac{{ab{c^2}}}{6}$
${a^2}bc = a{b^2}c = ab{c^2}$
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2 we obtain ${a^2}bc = a{b^2}c$ and $a{b^2}c = ab{c^2}$.
From the first and the second equation we obtain $a=b$ and $b=c$, respectively. Thus, the solution is $a=b=c$.
So, the constraint becomes
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1 = 0$
$\frac{3}{a} = 1$
So, $a=3$.
Since $a=b=c$, the critical point is $\left( {a,b,c} \right) = \left( {3,3,3} \right)$.
Step 4. Calculate the critical values
So, the extreme value of $V$ at $\left( {a,b,c} \right) = \left( {3,3,3} \right)$ is $V = \frac{9}{2}$.
Since the volume $V = \frac{1}{6}abc$ depends on $a$, $b$ and $c$, we can increase any edge of the tetrahedron as large as possible but at the same time adjust the other edges such that the plane still pass through the point $P = \left( {1,1,1} \right)$. This is illustrated in the figure attached. Hence, the volume can be as large as possible. Thus, we conclude that the extreme value $\frac{9}{2}$ is the minimum value of $V$ among all planes passing through the point $P = \left( {1,1,1} \right)$.
Notice that there is no maximum value since there are edges on a plane that are arbitrarily far from the origin.
