There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is:

Let us denote the number of men by the symbol $$m$$. Besides these men there are exactly two women in the tournament, so the total number of participants is $$m+2$$.

By the condition of the tournament, every distinct pair of participants plays two games with each other. We must compare two separate counts:

1. The total number of games played among the men themselves.

2. The total number of games played between the men and the women.

To find the first count, we start with the well-known combination formula. The number of unordered pairs that can be chosen from $$m$$ men is

$$\binom{m}{2}=\frac{m(m-1)}{2}.$$

For each such pair, exactly two games are played. Therefore, the number of games only among the men is

$$\text{Games among men}=2 \times \binom{m}{2}=2 \times \frac{m(m-1)}{2}=m(m-1).$$

Next, we consider the games between men and women. There are $$m$$ ways to pick a man and $$2$$ ways to pick a woman, giving $$m \times 2 = 2m$$ distinct man-woman pairs. Each such pair also plays two games, so

$$\text{Games between men and women}=2 \times (2m)=4m.$$

The problem states that the number of games among the men exceeds the number of games between men and women by $$84$$. Translating this directly into an equation, we have

$$m(m-1)-4m=84.$$

Now we simplify step by step:

First expand the left-hand side:

$$m(m-1)=m^2-m,$$

so

$$m^2-m-4m=84.$$

Combine the like terms $$-m-4m$$ to get $$-5m$$:

$$m^2-5m=84.$$

Bring all terms to one side to obtain a standard quadratic equation:

$$m^2-5m-84=0.$$

We now solve this quadratic by the quadratic formula. For an equation of the form $$ax^2+bx+c=0$$ the solutions are

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

Here we have $$a=1,\; b=-5,\; c=-84$$. Substituting these values, we get

$$m=\frac{-(-5)\pm\sqrt{(-5)^2-4\cdot1\cdot(-84)}}{2\cdot1}.$$

Simplify each part carefully:

$$-(-5)=5,$$

$$(-5)^2=25,$$

$$-4\cdot1\cdot(-84)=336,$$

so inside the square root we have

$$25+336=361.$$

Therefore,

$$m=\frac{5\pm\sqrt{361}}{2}.$$

Since $$\sqrt{361}=19$$, we obtain

$$m=\frac{5\pm19}{2}.$$

This gives two numerical possibilities:

$$m=\frac{5+19}{2}=24/2=12,$$

or

$$m=\frac{5-19}{2}=-14/2=-7.$$

The value $$m=-7$$ has no meaning in this context because a negative number of men is impossible. Hence the only permissible solution is

$$m=12.$$

Among the given options, $$12$$ corresponds to Option B.

Hence, the correct answer is Option B.