Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$?
$\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}$
[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations pair P,Q,R,Pp,Qp,Rp; pair A,B;  //Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); Pp=foot(P,A,B); Qp=foot(Q,A,B); Rp=foot(R,A,B); path PQR = P--Q--R--cycle; //Initial Diagram dot(P); dot(Q); dot(R); dot(Pp); dot(Qp); dot(Rp); draw(Circle(P, 1), linewidth(0.8)); draw(Circle(Q, 2), linewidth(0.8)); draw(Circle(R, 3), linewidth(0.8)); draw(A--B,Arrows); label("$P$",P,N); label("$Q$",Q,N); label("$R$",R,N); label("$P'$",Pp,S); label("$Q'$",Qp,S); label("$R'$",Rp,S); label("$l$",B,E);  //Added lines draw(PQR); draw(P--Pp); draw(Q--Qp); draw(R--Rp);  //Angle marks draw(rightanglemark(P,Pp,B)); draw(rightanglemark(Q,Qp,B)); draw(rightanglemark(R,Rp,B)); [/asy]
Notice that we can find $[P'PQRR']$ in two different ways: $[P'PQQ']+[Q'QRR']$ and $[PQR]+[P'PRR'].$ Since we want $[PQR],$ we use the latter method, so we have $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR'].$
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$P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$. Additionally, $Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}$. Therefore, $[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}$. Similarly, $[Q'QRR']=5\sqrt6$. We can calculate $[P'PRR']$ easily because $P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}$. $[P'PRR']=4\sqrt{2}+4\sqrt{6}$. $\newline$
Plugging into first equation, the two sums of areas, $3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR],$ so therefore $[PQR]=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}$