Elementary Set theory question .... it was asked in Exam There are $21$ people.
$9$ eat dish $A$
$10$ eat dish $B$
$7$ eat dish $C$
$5$ eat dish $A , B$ and $C$
How many people eat at least two dishes?
Answer:

$10$ (given in solutions)
$15$ (as per me )

Please tell me which one is correct. Also, tell if youu have different answer.

There seems to be something wrong here.
Consider this Venn diagram and its labeled parts.

We are told that: 
$$a+e+f+g=|A|=9 \qquad b+d+f+g=|B|=10 \qquad c+e+d+g=|C|= 7$$
$$g = 5$$
Thus
$$\begin{align}
a+e+f=4 \\
b+d+f=5 \\
c+e+d=2
\end{align}$$
so that, by adding all three equations,
$$a+b+c+2d+2e+2f=11 \qquad(1)$$
But, assuming each of the 21 people eats at least one dish, we also know that
$$a+b+c+d+e+f+g=21$$
so that (since $g=5$)
$$a+b+c+d+e+f=16 \qquad(2)$$
Consequently, by subtracting equation (2) from equation (1),
$$d+e+f = -5 \text{ (!)}$$
This problematic result is consistent with @Brian's work, which concludes
$$\begin{align}
|A\cap B| + |A\cap C|+|B\cap C| &= (f+g)+(e+g)+(d+g) \\
&= 3g+(d+e+f) \\
&= 3\cdot 5 + (-5) \\ 
&= 10 \\
\end{align}$$
Note, though, that this value does not answer OP's question, as it over-counts the people eating all three dishes. The number of people eating at least two dishes should be given by $d+e+f+g$ (that is, $|A\cap B|+|A \cap C|+|B\cap C|-2|A\cap B\cap C|$), but the computed value here is zero. Weird.

Edit. Perhaps the assumption "each of the 21 people eats at least one dish" is in error. Let $h$ be the number of people who eat nothing. Then we have
$$a+b+c+d+e+f+g+h=21$$
and
$$d+e+f=h-5$$
so that (barring negative people from the dinner party) $h \ge 5$. Moreover,
$$21 = a+b+c+(h-5)+5+h = a+b+c+2h$$
Therefore (also barring fractional people from the dinner party), $h \le 10$, and we can write
$$5 \le d+e+f+g \le 10$$
I don't see the conditions that force us to accept $10$ for the value of $d+e+f+g$; indeed, I've found scenarios $(a,b,c,d,e,f,g,h)$ that give rise to each possible value of the expression:
$$\begin{align}
(0,0,1,1,0,4,5,10) \quad &\implies \quad d+e+f+g = h = 10 \\
(1,2,0,1,1,2,5,9) \quad &\implies \quad d+e+f+g = h = 9 \\
(1,3,1,0,1,2,5,8) \quad &\implies \quad d+e+f+g = h = 8 \\
(3,4,0,1,1,0,5,7) \quad &\implies \quad d+e+f+g = h = 7 \\
(3,4,2,0,0,1,5,6) \quad &\implies \quad d+e+f+g = h = 6 \\
(4,5,2,0,0,0,5,5) \quad &\implies \quad d+e+f+g = h = 5
\end{align}$$
