答案：A
$\left\{ \begin{array}{ll}
3x – y – 2 =0 & \ldots \unicode{x2460} \\
5x^2 + 5y^2 +kx +4y -20 =0 & \ldots \unicode{x2461}
\end{array} \right.$
$\left\{ \begin{array}{ll}
3x – y – 2 =0 & \ldots \unicode{x2460} \\
5x^2 + 5y^2 +kx +4y -20 =0 & \ldots \unicode{x2461}
\end{array} \right.$
從 $\unicode{x2460}$，可得
$\begin{array}{rcl}
3x – y – 2 & = & 0 \\
y & = & 3x – 2
\end{array}$
把 $y = 3x – 2$ 代入 $\unicode{x2461}$，可得
$\begin{array}{rcl}
5x^2 + 5(3x – 2)^2 + kx + 4(3x – 2) – 20 & = & 0 \\
5x^2 + 45x^2 -60x + 20 + kx + 12x – 8 – 20 & = & 0 \\
50x^2 + (k – 48)x – 8 & = & 0
\end{array}$
由於 $PQ$ 的中點的 $x$ 坐標為 $2$，可得
$\begin{array}{rcl}
\dfrac{\text{兩根之和}}{2} & = & 2 \\
\dfrac{-(k -48)}{50} \times \dfrac{1}{2} & = & 2 \\
-(k – 48) & = & 200 \\
k – 48 & = & -200 \\
k & = & -152
\end{array}$
