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?

BTW: you don't need Lagrange multipliers for this problem:
To find the minimum of $abc$ for $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$ you can use $AM-GM$:
$$\frac{1}{3}=\frac{\frac{1}{a}+\frac{2}{b}+\frac{3}{c}}{3} \geq \sqrt[3]{\frac{6}{abc}}$$
Thus
$$abc \geq 3^3 \cdot 6 \,,$$
with equality if and only if 
$$\frac{1}{a}=\frac{2}{b}=\frac{3}{c}$$
