2016 AMC 10A Problem 21

Put the tangent line on the $x$-axis and take $P=(0,1).$ Because $PQ=1+2=3$ and the centers differ in height by $1,$ the horizontal distance from $P$ to $Q$ is $$\sqrt{3^2-1^2}=2\sqrt2.$$ Similarly, $QR=2+3=5$ and its centers also differ in height by $1,$ so their horizontal distance is $2\sqrt6.$ Thus we may use $$\begin{aligned} P&=(0,1),\qquad Q=(2\sqrt2,2), \\ R&=(2\sqrt2+2\sqrt6,3). \end{aligned}$$ The coordinate-area formula gives $$\begin{aligned} [PQR] &=\frac12\left|4\sqrt2-(2\sqrt2+2\sqrt6)\right| \\ &=\sqrt6-\sqrt2. \end{aligned}$$ Thus, the correct answer is D .