Let us
We can exchange the condition $\mathop{\rm vol}K=1$ to ${\rm vol}K_{\lbrace 1 \rbrace}=1$. Then In this case we need to show that $$\mathop{\rm vol}K\leqslant C\cdot \mathop{\rm vol}K_{\lbrace 1,2 \rbrace} \cdot \mathop{\rm vol}K_{\lbrace 1,3 \rbrace}.$$ The later boils down is equivalent to the following: $$\int\limits_0^1 u{\cdot}v{\cdot}dx\le u{\cdot}v{\cdot}dx\leqslant C\cdot \int\limits_0^1 u{\cdot}dx\cdot\int\limits_0^1 v{\cdot}dx\ \ \ \ \ (*)$$ for two positive convave functions $u,v\colon[0,1]\to\mathbb R$. WLOG we can assume that both functions are PL and both bave $0$ at $0$ andat $1$. Further we can assume that $$\int\limits_0^1 u{\cdot}dx=\int\limits_0^1 v{\cdot}dx=1. \ \ \ \ \ (**)$$ One We can move extremal points of $u$ and $v$ one by one keeping the identity $( * * )$ and increasing LHS of $( * )$. This way you we may reduce number of extremal points in $u$ and $v$; the process might terminate if you get one or two extramal extremal point in each.
It should be easy to check thenseems that $$u=v=2-4\cdot|x-\tfrac 12|$$ is the worst case. (I did not check it.)So $C=\tfrac43$ is the best constant.
