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Let $A(-1, 1)$, $B(3, 4)$ and $C(2, 0)$ be given three points. A line $y = mx$, $m > 0$, intersects lines $AC$ and $BC$ at point $P$ and $Q$ respectively. Let $A_1$ and $A_2$ be the areas of $\triangle ABC$ and $\triangle PQC$ respectively, such that $A_1 = 3A_2$, then the value of $m$ is equal to:
(1) $\frac{4}{15}$
(2) $1$
(3) $2$
(4) $3$
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