Plane passing through a given point so that the tetrahedron is of the minimum possible positive volume 
Find the plane passing through the point $(1,1,2)$ s. t. the volume of the tetrahedron enclosed by the plane and the coordinate axes is of the minimum possible positive volume.

$$A(x-1)+B(y-1)+C(z-2)=0\\Ax+By+Cz=A+B+2C\tag 1$$ the plane shouldn't be parallel to any of the axes so $A,B,C\ne 0.$ Plugging $(x_0,0,0),(0,y_0,0),(0,0,z_0)$ into $(1),$ I got: $$\begin{aligned}x_0&=1+\frac{B}A+\frac{2C}A\\y_0&=\frac{A}B+1+\frac{2C}B\\z_0&=\frac{A}C+\frac{B}C+2\end{aligned}$$
Then, I tried to find the minimum of the function  $\begin{aligned}f(x,y,z)=\left(1+\frac{y}x+\frac{2z}x\right)\left(\frac{x}y+1+\frac{2z}y\right)\left(\frac{x}z+\frac{y}z+2\right)\\=\frac{(x+y+2z)^3}{xyz}\end{aligned}$
$x,y,z$ should be of the same sign.
$$\frac{(x+y+2z)^3}{xyz}\ge\frac{(3\sqrt[3]{2xyz})^3}{xyz}=27\cdot 2=54,$$ which is attained when $x=y=2z.$
So, my answer is $$\pi\ldots 2(x-1)+2(y-1)+z-2=0$$
Can somebody verify my answer?
 A: Let the normal to the plane be $N = (1, A, B) $, then the equation of the plane is
$N \cdot (r - (1, 1, 2) ) = 0 $
which simplifies to
$ (x - 1) + A (y - 1) + B (z - 2) = 0 $
Setting $y=z=0$, gives us
$ a = 1 + A + 2 B $
Setting $x = z = 0 $ gives us
$ b = 1 + \dfrac{1}{A} ( 1 + 2 B ) $
Setting $ x = y = 0$ gives us
$ c = 2 + \dfrac{1}{B} (1 + A) $
Then we want the product $abc$ to be minimum, i.e.
$abc = (1 + A + 2 B) (1 + \dfrac{1 + 2 B}{A} ) (2 + \dfrac{1+A}{B} )$
This simplifies to
$ abc = f(A, B) =  \dfrac{ (1 + A + 2 B)^3}{AB} $
$ f_A = \dfrac{ 3(1 + A + 2 B)^2 (A B) - B (1 + A + 2 B)^3 }{A^2 B^2} $
$f_B = \dfrac{ 3 (1 + A + 2 B)^2 (2 A B) - A (1 + A + 2 B)^3 }{A^2 B^2} $
Therefore, at the critical point, $A , B$ satisfy
$  3 A - (1 + A + 2 B) = 0 $
$ 6 B - (1 + A + 2 B) = 0 $
whose solution is $ A = 1 , B = \dfrac{1}{2} $
Hence,
$ a = 3 , b = 3, c = 6 $
Which gives the minimum volume of the tetrahedron as $ \dfrac{1}{6} (3)(3)(6) = 9 $
And the equation of the plane is
$ (x - 1) + (1) (y - 1) + \dfrac{1}{2} ( z - 2) = 0 $
which simplifies to
$ 2 (x - 1) + 2 (y - 1) + z - 2 = 0 $
So your answer is correct.
