A solid tetrahedron is sliced off a wooden rectangular solid measuring $1\times2\times3$ by the plane through the three vertices adjacent to one vertex $A$ of the solid. The tetrahedron is discarded, and the remaining solid is placed on a table with the cut face down. What is the height of this solid?

1. Put $A=(0,0,0)$ and the solid as $[0,1]\times[0,2]\times[0,3]$. The three vertices adjacent to $A$ are $(1,0,0),(0,2,0),(0,0,3)$, so the cutting plane is $\frac x1+\frac y2+\frac z3=1$, i.e. $6x+3y+2z=6$.
2. The remaining solid lies on the side $6x+3y+2z\ge6$. Its highest point above the table (the cut plane) is its vertex farthest from the plane, namely the vertex opposite $A$, $(1,2,3)$ (the other vertices are closer: e.g. $(1,2,0)$ is at distance $\frac67$).
3. The distance from $(1,2,3)$ to the plane is $\dfrac{6\cdot1+3\cdot2+2\cdot3-6}{\sqrt{6^2+3^2+2^2}}=\dfrac{12}{\sqrt{49}}=\boxed{\dfrac{12}7}$.