The problem: Let $\Bbb R_+^3$ be the first coordinate octant in $R^3$. Then for each fixed $\alpha > 0$ describe the geometric location of points $A \in \Bbb R_+^3$, which satisfy the following condition: the smallest volume of the part of the octant cut off by an arbitrary plane drawn through $A$ is equal to $\alpha$.
As I understand I need to find points. Part of octant will be a pyramid. Its volume must be equal to $\alpha$. We can use the formula: $$V = \frac{1}{3}S \cdot h$$ So, $$\alpha = \frac{1}{3}S \cdot h$$ As there must be at least $3$ points to draw the plane, then they will be at $(x;0;0)$, $(0;y;0)$, $(0;0;z)$. So I can rewrite our formula that way: $$\alpha = \frac{1}{3} \cdot \frac{1}{2} \cdot x \cdot y \cdot z = \frac{xyz}{6} $$ But it's not the full answer. I don't know what to do next. Tell me if I'm wrong somewhere and what should I do?
Let $P=(x,y,z)$ be any point in the first octant. Any plane cutting the positive semi-axes at $(a,0,0)$, $(0,b,0)$, $(0,0,c)$ passes through $P$ if: $$ {x\over a}+{y\over b}+{z\over c}=1 $$ and the volume cut off by such a plane is: $$ V={1\over 6}abc. $$ By AM-GM inequality we have: $$ \root{3}\of{1\over V}=\root{3}\of{6\over abc}= \root{3}\of{6\over xyz}\root{3}\of{xyz\over abc}\le \root{3}\of{6\over xyz}{1\over3}\left({x\over a}+{y\over b}+{z\over c}\right)= {1\over3}\root{3}\of{6\over xyz}$$ with equality reached when $$ {x\over a}={y\over b}={z\over c}={1\over3}. $$ Hence the minimum volume is $$ \alpha=27{xyz\over 6} $$ and this is the equation of the requested locus.