$A\triangle B$ can be defined either as $(A\cup B)\setminus(A\cap B)$ or as $(A\setminus B)\cup(B\setminus A)$. The two definitions are entirely equivalent, but it’s the latter that explains why $A\triangle B$ is called the symmetric difference of $A$ and $B$: $A\setminus B$ is the difference in one direction, $B\setminus A$ is the difference in the other direction, and taking their union removes any directionality. From an intuitive point of view, however, you might be best off thinking of $A\triangle B$ simply as the set of things belonging to exactly one of $A$ and $B$, just as $A\cap B$ is the set of things belonging to exactly two of $A$ and $B$, and $A\cup B$ is the set of things belonging to at least one of $A$ and $B$.
Now let’s look at $A\triangle(B\cap C)$. $A$ is striped red in the figure below, and $B\cap C$ is solid blue. The points that are in exactly one of those two sets are exactly the points shaded in your picture.
And here they are again, in a modified version of my picture: the remaining blue points are the points that are in $B\cap C$ but not in $A$ (in symbols, in $(B\cap C)\setminus A$), and the remaining red-and-white striped region contains the points that are in $A$ but not in $B\cap C$ (in symbols, in $A\setminus(B\cap C)$). Between the two we have
$$A\triangle(B\cap C)=\underbrace{\Big(A\setminus(B\cap C)\Big)}_{\text{remaining striped region}}\cup\underbrace{\Big((B\cap C)\setminus A\Big)}_{\text{remaining blue region}}\;.$$
Alternatively, we started with the things that were in at least one of the sets $A$ and $B\cap C$ and removed the things that were in both to get the things in exactly one:
$$A\triangle(B\cap C)=\underbrace{\Big(A\cup(B\cap C)\Big)}_{\text{in at least one}}\setminus\underbrace{\Big(A\cap(B\cap C)\Big)}_{\text{in both}}\;.$$
The picture that you already have for $(A\triangle B)\cap(A\triangle C)$ is pretty good. The set $A\triangle B$ is shaded from upper left to lower right (bendwise, if you’re a herald), and the set $A\triangle C$ is shaded from upper right to lower left (bendwise sinister if you’re a herald). The intersection of these two sets consists of those points that are in both sets, so it comprises the regions that are shaded in both directions. In the picture below it’s the blue together with the red-and-white regions.
(The diagrams are a bit crude, but they should help a bit, at least.)
There are a couple of ways to approach the problem; I’ll start with a very straightforward one from first principles. First, here’s a picture:
The blue part of the diagram represents those students who do not participate in soccer, basketball, or track. We’re interested in $x$, the percentage who play all three sports. One way to solve the problem is to fill in the Venn diagram in terms of $x$. I’ve started: we know, for instance, that $2\%$ play soccer and basketball, and $x\%$ play all three sports, so $2-x$ must be the percentage playing soccer and basketball but not track. The figures $6-x$ and $9-x$ were obtained by similar reasoning. Now go after the percentage playing just soccer, just basketball, and just track. We know that $21\%$ play soccer, and $(2-x)+x+(6-x)$ percent play soccer and at least one other of the three sports, so what percentage (in terms of $x$) play just soccer? Repeat this procedure with basketball and track to fill in the last three regions of the Venn diagram. Finally, add up the percentages of all of the white regions; that total, which will involve $x$, must be the $100-36$ percent who play at least one of the three sports, and you’ll have a simple equation to solve for $x$.
If you know something about calculating the cardinalities of unions and intersections of sets, you can short-circuit this calculation. Let $S,B$, and $T$ be the sets of students playing soccer, basketball, and track, respectively. Pretend that there are $100$ students altogether, so that percentages are simply numbers of students. Then you want $|S\cap B\cap T|$, and you know $|S|$, $|B|$, $|T|$, $|S\cap B|$, $|S\cap T|$, $|B\cap T|$, and the cardinality of the complement of $S\cup B\cup T$. That last figure gives you $|S\cup B\cup T|$. Now, do you know a formula that relates all of these quantities? If so, you’re in business, because you know all but one of them.
Added: However, the problem is faulty, because if you actually carry out the computations, which I unfortunately did not bother to do when I originally posted the answer, you’ll find that $x$ is negative!
Best Answer
I think you're misinterpreting what the symmetric difference $A \oplus B$ is: it's all the points that are in exactly one of $A$ or $B$, but not both. In particular, this means that the "exclusively $C$" region is not included in $A \oplus B$, because elements exclusively in $C$ are in neither $A$ nor $B$.
Based on this, to fix your diagram you also need to shade the exclusively $C$ region.