Let us denote the ages of Person 1, Person 2, Person 3 by $x,y,z$ respectively. We'll assume that $x,y,z$ are positive throughout.
The product of the 1st person's and the 2nd person's ages is $311 \frac{2}{3}$ plus the 3rd person's age.
$$xy - z = 311 \frac{2}{3} = \frac{935}{3}$$
The sum of the 1st person's age and the quotient of the 3rd person's and the 2nd person's ages is $41 \frac{17}{24}$
$$x + \frac{z}{y} = 41 \frac{17}{24} = \frac{1001}{24}$$ $$\Rightarrow xy + z = \frac{1001}{24}y$$
Subtracting the first equation from the second gives
$$2z = \frac{1001}{24} y - \frac{935}{3} \Rightarrow z = \frac{1001}{48}y - \frac{935}{6}$$ $$\Rightarrow xy + \frac{1001}{48} y - \frac{935}{6} = \frac{1001}{24} y$$ $$\Rightarrow x = \frac{1001}{48} + \frac{935}{6y}$$
The difference between the 1st person's and the 2nd person's ages is $2 \frac{31}{33}$ times the 3rd person's age.
$$x-y = 2 \frac{31}{33} z = \frac{97}{33} z = \frac{97097}{1584}y - \frac{90695}{198} = \frac{8827}{144} y - \frac{8245}{18}$$ $$\Rightarrow x = \frac{8971}{144}y - \frac{8245}{18} = \frac{1001}{48} + \frac{935}{6y}$$ Multiplying across by $144y$ and rearranging gives $$\Rightarrow 8971 y^2 - 68963 y - 22440 = 0$$ Solving the quadratic equation for $y$ gives $$ y = \frac{68963 \pm \sqrt{4755895369 + 805236960}}{17942} = \frac{68963 \pm \sqrt{5561132329}}{17942} = \frac{68963 \pm 74573}{17942}$$ Note here that calculating the square root is the trickiest part of this whole calculation but if we notice that our given number is a little less than $56 \times 10^8$ then we should expect the result to be a little less than $7.5 \times 10^4$. Using some finer estimation like this allows us to hone in on the answer a little quicker and we can use a binary search to narrow down to the given value.
Using the assumption that $y$ be positive, we find that $$y = \frac{143536}{17942} = 8$$ and substituting into the equations for $x$ and $z$ we find that $$ x = \frac{1001}{48} + \frac{935}{48} = \frac{1936}{48} = 40 \frac{1}{3}\,\,\,\,,\,\,\,\, z = \frac{1001}{6} - \frac{935}{6} = \frac{66}{6} = 11$$
Now let us check the consistency with the other equations
The square of the 3rd person's age is triple the 1st person's age
$$3x = 3\left(\frac{121}{3}\right) = 121 = 11^2 = z^2$$ so this is consistent
The quotient of the product of all their ages and the sum of all their ages is $826 \frac{4}{29}$
A quick check is enough to convince us that $$ \frac{xyz}{x+y+z} \neq 826 \frac{4}{29}$$ However, the question does not specifically state that there are only $3$ people at the party so let us assume there is a fourth person, Person 4 whose age is $w$. Then we require $$ \frac{xyzw}{x+y+z+w} = \frac{10648w}{178 + 3w} = \frac{23958}{29}$$ $$\Rightarrow 308792w = 4264524 + 71874w$$ $$\Rightarrow w = \frac{4264524}{236918} = 18$$
Whose birthday is it?
Since theirs are exact ages it must be the birthdays of Person 2, Person 3 and Person 4 and they are $8, 11$ and $18$, respectively
