jmc

Let $A=(0,0,0),$ $C=(1,0,0),$ $Q=(0,0,1),$ and $R=(1,1,1).$ It is clear that the the shortest path is obtained by travelling from $A$ to some point $B$ directly on a line segment (where $B$ is some point on line segment $\overline{QR}$), then travelling from $B$ to $C$ on another line segment. The only question is then where to place point $B.$



Let $M$ be the midpoint of $\overline{QR},$ which would be $\left(\frac{1}{2},\frac{1}{2},1\right),$ and consider the circle centered at $M$ with radius $MC=\sqrt{\frac{3}{2}},$ contained in the plane that is perpendicular to line $\ell .$ Let $P$ be the "top" point of this circle, so $P=\left(\frac{1}{2},\frac{1}{2},1+\sqrt{\frac{3}{2}}\right).$ Note that right triangles $BMC$ and $BMP$ are congruent, so $BC=BP.$ This means $$AB+BC=AB+BP.$$Let $B^{\prime }$ be the intersection of $\overline{AP}$ with line $\ell .$ By the Triangle Inequality, $$AB+BP\ge AP.$$Equality occurs when $B$ coincides with $B^{\prime }.$ Thus, the minimum value of $AB+BP$ is $AP=\sqrt{3+\sqrt{6}},$ so the final answer is $A{P}^2=\boxed{3+\sqrt{6}}.$