The solution is the row-selector, trying to maximize the value V, chooses row 1 25% of the time and row 3 75% of the time. The column-selector, trying to minimize V, chooses column 1 half the time and column 3 half the time.
The value of the game (assuming the row selector is trying to maximize) is 5/2.
Your equations miss the point: the right equations to use are of the form (for example) of
$$
V \leq P_1 - P_2 + 3P_3
$$
$$
V \leq -P_1 +5 P_2 + 3P_3
$$
$$
V \leq 4P_1 +3 P_2 + 2P_3
$$
$$
P_1 + P_2 + P_3 = 1
$$
and maximize V subject to all variables $P_i \geq 0$.
For this specific game, you can immediately tell that the second column is dominated by the first, and after eliminating the second column, that the second row is dominated by the first, leaving a 2x2 game with the solution as given.
Although the general way to solve such games is the simplex method, and that actually is not too ugly to do for a 3x3 game matrix, my mantra when given a 3x3 game is:
Form, for the rows, two extra twos consisting of the top row minus the second, then top minus the third. Among those two rows, write the determinant of the (1,2) columns square matrix under the third column, the determinant of the (2,3) square under the first column, and the determinant of the (3,1) column square under the second column.
Then do the same for the columns, subtracting columns 2 and 3 from the first.
$$
\begin{array}{c c c | c c | c}
1 & 2 & 4 & -1 & -3 & -6 \\
-1 & 5 & 3 & -6 & -4 & +1 \\
3 & 3 & 2 & 0 & 1 & -14 \\
\hline
2 & -3 & 1 & & & \\
-2 & -1 & 2 & & & \\
\hline
-5 & -6 & -8 & & &
\end{array}
$$
If the answers for the rows are all the same sign, and the answers for the columns are all the same sign, then the game solution requires use of all three rows and all three columns, and you read off the strategies as the ratios of the numbers you just found. But in this case, that does not work because the column numbers are -6, +1, and -14. That means that the correct solution will use fewer than three rows.
Try eliminating one row and solving the resulting 2x3 game: First you might try eliminating row 1. But the solution to that game (of value 11/5) has the column chooser selecting column 3 80% of the time and column 1 20%; clearly the row chooser can do better than 11/5 by picking row 1.
When we eliminate row 2, we get the right solution, which is the solution presented above.
It is easy to check that neither player can improve his result by choosing one of the eliminated choices.
In some rare cases, you have a situation where there are a family of correct solutions, but each member of the family requires a mix of all 3 rows (or columns) while the strategy for the other player is always the same ratio of only two of the columns (or rows). This makes the solution less obvious, although the simplex method always works. I may present such a game as a problem on this site at a later date.
Best Answer
Given a payoff matrix of size n, the gain of a mixed strategy $(p_1, p_2, ... p_n)$ can be expressed as a polynomial (quadratic) function of $(p_1, p_2, ... p_{n-1})$ in the polytope defined by $0\le p_i\le 1$ and $\sum_{1\le i<n}p_i\le 1$.
Finding the optimal strategies is related to find all local maximum of the function : in the general case, you need to evalute the gradient and the eigenvalues of the Hessian matrix (you can read more at https://en.wikipedia.org/wiki/Hessian_matrix#Critical_points)
If you find one maximum inside the polytope, then you have your mixed strategy (there can by only one as the function is quadratic).
If it fails (no maximum inside the polytope) then you need to find maximum on the borders (each part of the border is defined by $p_i=0$ for some $1\le i\le n$) by applying the same method with one less variable. Each border can have its own set of local maximum.
There is always the possibility to have degenerate solution (with eigenvalues = 0) that can imply that the set of optimal strategies is not a set of isolated points (but something larger, like an hyperplan $ p_1=p_2$).
In your example : $$g(p_1,p_2)=2p_1p_2+4p_1p_3+6p_2p_3 $$ by replacing $p_3=1-p_1-p_2$ $$g(p_1,p_2)=-4p_1^2-8p_1p_2-6p_2^2+4p_1+6p_2$$
the gradient is (first derivatives) $(-8p_1-8p_2+4, -8p_1-12p_2+6)$ that is zero iff $(p_1=0, p_2=\frac{1}{2})$. The Hessian matrix is a constant matrix (the second derivatives of a quadratic function) and here it's $$\left[\begin{matrix} -8 & -8 \\ -8 & -12\\ \end{matrix}\right]$$ The eigenvalues are all negative (as the determinant is positive (32) and the trace is negative (-20)), so it's a global maximum at $(0,\frac{1}{2},\frac{1}{2})$. No need to look at the borders (which can be quite complex).
So there is only one stable and optimal strategy for this game.