$\quad$ Let $x_k, y_k, z_k$ the number of apples, pears, plums after the $k$-th trick. $x_0=y_0=z_0=2012$. After a trick $i)$, $x_{k+1}=x_k+2, y_{k+1}=y_k-1, z_{k+1}=z_k-1$. After a trick $ii)$, $x_{k+1}=x_k-1, y_{k+1}=y_k-1, z_{k+1}=z_k+3$. After a trick $iii)$, $x_{k+1}=x_k-1, y_{k+1}=y_k+4, z_{k+1}=z_k-1$. $\quad$ After $m$ tricks $i)$, $n$ tricks $ii)$ and $p$ tricks $iii)$, if $m+n+p=s$, we obtain: $\begin{cases} x_s-x_0=2m-n-p=0\;(1);\\y_s-y_0=-m-n+4p=0\;(2);\\z_s-z_0=-m+3n-p>0\;(3).\end{cases}$. From $(1)$ results: $p=2m-n$; From $(2)$ results: $4p=m+n=8m-4n\Longrightarrow 7m=5n\Longrightarrow$ $\Longrightarrow m=5k, n=7k, p=3k, k\in\mathbb{N}$. The number of plums becomes $z_s=2012+(-m+3n-p)=2012+13k$. The minimal number of plums is $\min{z_s}=z_{15}=2012+13=2025$, after $5$ tricks $i)$, $7$ tricks $ii)$ and $3$ tricks $iii)$.