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.
I am still not sure if I really got your point, however, I shall try to provide at least a small example -- but not a proof -- to give some evidence that every element of the pre-kernel also satisfies the standard solution for every two person DM-reduced game. For a proof of RGP and the pre-kernel -- that includes this special case -- have a look at Peleg and Sudhoelter (2007).
For this purpose, let us resume the game I have presented at
Four Person Game
I consider here only the reduced game of $T:=\{2,4\}$ w.r.t. the pre-kernel element $\mathbf{x}:=\{1,3,1,3\}/4$. This game has coalitional values of
$$v_{\mathbf{x},T}(\{2\})= 3/4 \qquad v_{\mathbf{x},T}(\{4\}) = 3/4 \qquad v_{\mathbf{x},T}(\{2,4\}) = 3/2, $$
which is a two person DM-reduced game. Applying now the standard solution, this gives $\vec{y}=\{3,3\}/4$. However, one can check out that the pre-kernel is given by $\mathcal{PK}(T,v_{\mathbf{x},T})=\{3,3\}/4$, which coincides with the pre-nucleolus. Since up to three person games both solution concepts are identical. Thus, we get $$\{\mathbf{x}_{T}\} = \mathcal{PK}(T,v_{\mathbf{x},T}) = \mathcal{PN}(T,v_{\mathbf{x},T}).$$
This examples reproduces the theoretical expected result that whenever $\langle N,v \rangle \in \Gamma, \emptyset \neq S \subseteq N$, and $\mathbf{x} \in \mathcal{PK}(N,v)$, then $\langle N,v_{\mathbf{x},S} \rangle \in \Gamma$ and $\mathbf{x}_{S} \in \mathcal{PK}(S,v_{\mathbf{x},S})$ must hold. This also includes the standard solution for every $\arrowvert S \arrowvert = 2$.
Best Answer
In a sense, that is the definition. Or to be more precise, the indifference principle states that:
Then for this specific game, since the expected payoff for guessing $G$ is $G\times Prob(G=n) = G \times p_G$, the indifference principle becomes the formula $1 p_1 = 2 p_2 = 3 p_3 = \cdots$
So that answers your question: Yes it is the definition, or rather, a direct, immediate consequence of the definition.
The underlying question, which you didn't ask, is: Why should the picker adopt the indifference principle? I.e., why is indifference the "optimal" strategy?
If the picker has some insight into the personal tendencies of the guesser, then the picker might do something different to foil the guesser. E.g. if the picker knows the guesser is bad at math and will most likely guess $100$ thinking all numbers are equally likely and that number pays the most when correct, then the picker can foil the guesser by picking $1$. But of course, the guesser, knowing that, would have picked $1$, and the picker, knowing that, might have picked $2$ instead, etc. This turns into a game of think and double-think.
The indifference principle avoids all this. Essentially, the picker is assuming the worst case "meta" scenario: that the guesser knows the picker's probability numbers $p_i$ for all $i$. Perhaps the guesser has a spy, or can read mind, or is just really really good at math and logic. Under this worst case scenario, the best (optimal) the picker can do is to adopt the indifference principle. If the picker did anything else, the guesser (knowing all $p_i$ values) can foil it and obtain a higher expected payoff by guessing $G= \arg\max_i i p_i$.
The indifference principle is even easier to explain in Rock-Paper-Scissors. One can certainly play it as think-vs-double-think, but most people(?) try to adopt the indifference principle $p_R=p_P=p_S = 1/3$ s.t. no opponent can have any extra benefit.
Note that the indifference principle does give up something: in return for protection against a really good opponent, you give up the opportunity to exploit a really bad opponent. E.g. if your opponent always plays Rock (or always guesses $100$), your indifference principle would still give the same expected payoff - and in a sense, let your opponent "off the hook". In other words, indifference is optimal under the "worst case meta scenario" I described above, but it may not be optimal if you know in advance your opponent's (probabilistic) strategy.
You can read much more about this in the context of Nash equilibrium