Find an equation of the plane that passes through the point $(1,2,3)$, and cuts off the smallest volume in the first octant. *help needed please* Find an equation of the plane that passes through the point $(1,2,3)$, and cuts off the smallest volume in the first octant.
This is what i've done so far....
Let $a,b,c$ be some points that the plane cuts the $x,y,z$ axes. --> $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a,b,c >0$. 
I saw a solution for this question was to use Lagrange multiplier. The solution goes as follows... 
The product $abc$ will be equal to $6$ times the volume of the tetrahedron $OABC$ (could someone explain to my why is this so?)
$f(a,b,c) = abc$ given the condition $(\frac1a + \frac2b + \frac3b -1)$
$f(a,b,c) = abc + \lambda (\frac1a + \frac2b + \frac3c -1)$
2nd query to the question...
$f_a = \lambda g_a \Rightarrow  bc - \frac\lambda {a^2} ; a = \sqrt \frac \lambda {bc} 
\\f_b = \lambda g_b  \Rightarrow  ac - \frac\lambda {b^2} ; b = \sqrt \frac {2\lambda}{ac}
\\f_c = \lambda g_c  \Rightarrow  ab - \frac\lambda {c^2} ; c = \sqrt \frac {3\lambda}{ab}$
using values of $a,b,c$ into $\frac1a+\frac1b+\frac1c = 1\Rightarrow \lambda =\frac{abc}{a+2b+3c}$.
May i know how should i proceed to solve the unknowns?

The volume of a pyramid (of any shaped base) is $\frac13A_bh$, where $A_b$ is the area of the base and $h$ is the height (perpendicular distance from the base to the opposing vertex). In this particular case, we're considering a triangular pyramid, with the right triangle $OAB$ as a base and opposing vertex $C$. The area of the base is $\frac12ab$, and the height is $c$, so the volume of the tetrahedron is $\frac16abc$--equivalently, $abc$ is $6$ times the volume of the tetrahedron.
