Do the three ways of measuring producer’s surplus give the same result?
Ask your teacher what he meant, because there is either a misunderstanding or he is mistaken. 

Producer's surplus will come down to $y \cdot p - VC(y)$.

In the first graph, by definition of $AVC$ we have
$$
y \cdot (p - AVC(y)) = y \cdot p - VC(y).
$$
In the second graph, using $\int MC = VC$ and $VC(0) = 0$, we have
$$
\int_0^y p - MC(x) \text{d} x =
\left[x \cdot p - VC(x)\right]_0^y = y \cdot p - VC(y).
$$
The area in the third graph is 
$$
\begin{align*}
\int_0^z p - AVC(z) \text{d} x + 
\int_z^y p - MC(x) \text{d} x.
\end{align*}
$$
Let us calculate it by parts:
$$
\begin{align*}
\int_0^z p - AVC(z) \text{d} x &amp; = z \cdot p - z \cdot AVC(z) = z \cdot p -  VC(z) \\
 \\
\int_z^y p - MC(x) \text{d} x &amp; = (y - z) \cdot p - VC(y) + VC(z). 
\end{align*}
$$
Summing up the two, we get $y \cdot p - VC(y)$.