7. If four lines are drawn through the points $P(1,0), Q(2,0), R(4,0)$, $S(8,0)$ to form a square, then the area of the square cannot be equal to ( ).
(A) $\frac{16}{17}$
(B) $\frac{36}{5}$
(C) $\frac{26}{5}$
(D) $\frac{196}{53}$
7. C.

Assume the square formed by the intersection of the four lines is in the first quadrant, with side length $a$ and area $S$. Let the inclination angle of the line passing through point $P$ be $\theta\left(0<\theta<\frac{\pi}{2}\right)$.
When the lines passing through points $P$ and $Q$ are the lines of the opposite sides of the square,
$$
\begin{array}{l}
|P Q| \sin \theta=a=|R S| \cos \theta \\
\Leftrightarrow \sin \theta=4 \cos \theta \Leftrightarrow \sin \theta=\frac{4}{\sqrt{17}} .
\end{array}
$$

At this time, the area of the square
$$
S=(|P Q| \sin \theta)^{2}=\frac{16}{17} \text {. }
$$

Similarly, when the lines passing through points $P$ and $R$ are the lines of the opposite sides of the square, $S=\frac{36}{5}$;

When the lines passing through points $P$ and $S$ are the lines of the opposite sides of the square, $S=\frac{196}{53}$.