Answer: C

Small Hint:

Let $\theta$ be the angle line $PQ$ makes with the $x$-axis; the square condition relates the projections of the given segments

Big Hint:

From $2\cos\theta=6\sin\theta$ get $\tan\theta=\tfrac13,$ then intersect the two lines through the midpoints $(4,0)$ and $(10,0)$

Solution:

Let $\theta$ be the acute angle line $PQ$ makes with the $x$-axis. Sides $SR=PQ$ span the segment from $(3,0)$ to $(5,0)$ as $2\cos\theta,$ while $SP=QR$ span the segment from $(7,0)$ to $(13,0)$ as $6\sin\theta.$

Since the square has equal sides, $2\cos\theta=6\sin\theta,$ so $\tan\theta=\tfrac13.$ Thus lines $SP,RQ$ have slope $3$ and lines $SR,PQ$ have slope $-\tfrac13.$

The center lies on the line through $(4,0)$ with slope $3$ and the line through $(10,0)$ with slope $-\tfrac13:$ $$y=3(x-4),$$ $$y=-\tfrac13(x-10).$$ These meet at $(4.6,1.8).$

The sum of the coordinates is $4.6+1.8=6.4.$

Thus, the correct answer is C.