Let $A$, $B$, and $C$ be the set of pupils that have access to a PC running Windows, Apple, and Linux, respectively.
We know that
\begin{align*}
|A \cup B \cup C| &= 120 \\
|A| &= 80 \\
|B| &= 40 \\
|C| &= 10
\end{align*}
As @Tom van der Zanden pointed already pointed out in the comments, the wording of "... running the Apple or Windows operating system" is unclear. Thus I will cover all three interpretations of that wording.
Interpretation 1: "... running the Apple or Windows operating system" means the PC has installed both the Apple and Windows OS.
This means
\begin{align*}
|A \cap B| &= 30 \\
|A \cap C| &= 7 \\
|B \cap C| &= 5
\end{align*}
Using inclusion-exclusion, we come to the result that
\begin{align*}
|A \cap B \cap C| &= |A \cup B \cup C| - |A| - |B| - |C| + |A \cap B| + |A \cap C| + |B \cap C|\\
&= 120 - 80 - 40 - 10 + 30 + 7 + 5 \\
&= 32
\end{align*}
So your calculation is correct and there is a contraction. That means that either
- the inclusion-exclusion princliple does not hold in general, or
- the exercise description is inconsitent, i.e. some of the cardinalities are wrong.
Since the inclusion-exclusion principle is proven to hold for any arbitrary set of sets, the latter case must be true.
Interpretation 2: "... running the Apple or Windows operating system" means that the PC runs either the Apple or the Windows OS.
We can calculate the cardinalities of the intersections $A \cap B$, $A \cap C$, and $B \cap C$. However, we also run into a contradiction by using this interpretation:
\begin{align*}
& |(A \cup C) \setminus (A \cap C)| = |A \cup C| - |A \cap C| = |A| + |C| - 2|A \cap C| \\
\implies & |A \cap C| = \frac{|(A \cup C) \setminus (A \cap C)| - |A| - |C|}{-2} = \frac{7 - 80 - 10}{-2} = 41.5
\end{align*}
The above is a contradiction, since the cardinality of a finite set must be a natural number.
Interpretation 3: "... running the Apple or Windows operating system" is a regular logic or.
We have $|A| = 80 > 30 = |A \cup B|$. This is also a contradiction, since the cardinality of a set cannot be greater than the cardinality of its superset.
For either interpretation we show that there is a contradiction, therefore there must be a problem in the exercise description