jmc

We can write $$\begin{array}{rl}{A}^2-B^2& =(A+B)(A-B)\\ & =(\sqrt{x+2}+\sqrt{x+1}+\sqrt{y+5}+\sqrt{y+1}+\sqrt{z+10}+\sqrt{z+1})\\ & \times (\sqrt{x+2}-\sqrt{x+1}+\sqrt{y+5}-\sqrt{y+1}+\sqrt{z+10}-\sqrt{z+1}).\end{array}$$Let $$\begin{array}{rl}{a}_1& =\sqrt{x+2}+\sqrt{x+1},\\ b_1& =\sqrt{y+5}+\sqrt{y+1},\\ c_1& =\sqrt{z+10}+\sqrt{z+1},\\ a_2& =\sqrt{x+2}-\sqrt{x+1},\\ b_2& =\sqrt{y+5}-\sqrt{y+1},\\ c_2& =\sqrt{z+10}-\sqrt{z+1}.\end{array}$$Then by Cauchy-Schwarz, $$\begin{array}{rl}{A}^2-B^2& =(a_1+b_1+c_1)(a_2+b_2+c_2)\\ & \ge (\sqrt{a_1{a}_2}+\sqrt{b_1{b}_2}+\sqrt{c_2{c}_2}{)}^2\\ & =(1+2+3{)}^2\\ & =36.\end{array}$$Equality occurs when $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2},$$or equivalently, $$\frac{x+2}{x+1}=\frac{y+5}{y+1}=\frac{z+10}{z+1}.$$For example, if we set each fraction to 2, then we get $x=0,$ $y=3,$ and $z=8.$

Hence, the minimum value is $\boxed{36}.$