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
Let $(E)$ be the number of people with only blacked-eye, $(EW)$ with only blacked-eye and sprained wrist, $(EWT)$ with all three, so on.
We have:
$$\begin{align} (E)+(EW)+(ET)+(EWT)=\frac{2}{3}x \qquad(1)\\ (W)+(EW)+(WT)+(EWT)=\frac{3}{4}x\qquad(2)\\ (T)+(ET)+(WT)+(EWT)=\frac{4}{5}x\qquad(3)\\ (E)+(W)+(T)+(EW)+(ET)+(WT)+(EWT)=x\;\;\,\qquad(4)\\ \text{Minimum of }(EWT)=26\;\qquad(5) \end{align}$$
$$(1)+(2)+(3)-(4): \qquad (EW)+(ET)+(WT)+2(EWT)=\frac{73}{60}x$$
Here, the RHS is fixed, so in order for $(EWT)$ to have its minimum value, $(EW)+(ET)+(WT)$ has to have its maximum value. In $(4)$, $(E),(W),(T) \ge 0$ so the maximum is $x-(EWT)$.
$$\frac{73}{60}x=(EW)+(ET)+(WT)+2(EWT) \le x+(EWT)$$ $$\implies (EWT) \ge\frac{13}{60}x$$ $$\implies x=120$$