If $N$ denotes the total number of ways $x_1+x_2+x_3 =15$ without any restrictions (of course, non-negative integers) and $M$ is the number of ways with desired constraints, then
$$
M = N - \left(N_{x_1>5} + N_{x_2>5} + N_{x_3>7}\right) + \left(N_{x_1>5\ and\ x_2>5} + N_{x_1>5\ and\ x_3>7} + N_{x_2>5\ and\ x_3>7}\right) - N_{x_1>5\ and\ x_2>5\ and\ x_3>7}
$$
where the $N$ with a subscript denotes the number of ways with the constraints in the subscript.
We have $$N_{x_1>5} = N_{x_2>5} = {11\choose 2},\ N_{x_3>7} = {9\choose2},$$
$$N_{x_1>5\ and\ x_2>5} = {5\choose 2},\ N_{x_1>5\ and\ x_3>7} = N_{x_2>5\ and\ x_3>7} = {3\choose 2},$$ and $$N_{x_1>5\ and\ x_2>5\ and\ x_3>7} = 0.$$
And the total number of ways without any constraint is ${17\choose 2}$. So,
$$
M = {17\choose 2} - \left( 2{11\choose 2} + {9\choose2}\right) + \left({5\choose 2} + 2{3\choose 2}\right) = 6.
$$
I am not sure why you have four variables in the statement of your question and three variables in your answer. I will assume you meant to work with four variables.
How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_i > 0$ for each $i \in \{1, 2, 3, 4\}$?
We wish to solve the equation
$$x_1 + x_2 + x_3 + x_4 = 12 \tag{1}$$
in the positive integers.
Method 1: We reduce the problem to one in the non-negative integers. Let $y_k = x_k - 1$ for $1 \leq k \leq 4$. Then each $y_k$ is a non-negative integer. Substituting $y_k + 1$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields
\begin{align*}
y_1 + 1 + y_2 + 1 + y_3 + 1 + y_4 + 1 & = 12\\
y_1 + y_2 + y_3 + y_4 & = 8 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers. A particular solution corresponds to placing three addition signs in a row of eight ones. For instance,
$$1 1 1 1 1 + 1 + 1 1 1$$
corresponds to the solution $y_1 = 5$, $y_2 = 1$, and $y_3 = 3$, while
$$1 1 + + 1 1 1 1 1 1$$
corresponds to the solution $y_1 = 2$, $y_2 = 0$, and $y_3 = 6$. Thus, the number of solutions of equation 2 is the number of ways three addition signs can be inserted into a row of eight ones, which is
$$\binom{8 + 3}{3} = \binom{11}{3}$$
since we must choose which three of the eleven symbols (eight ones and three addition signs) will be addition signs.
Method 2: A particular solution of equation 1 in the positive integers corresponds to inserting three addition signs in the eleven spaces between successive ones in a row of $12$ ones.
$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance,
$$1 1 1 + 1 1 1 1 + 1 1 1 1 1$$
corresponds to the solution $x_1 = 3$, $x_2 = 4$, and $x_3 = 5$. Thus, the number of solutions of equation 1 in the positive integers is the number of ways three addition signs can be inserted into the eleven gaps between successive ones in a row of $12$ ones, which is
$$\binom{11}{3}$$
How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \geq 0$?
We reduce the problem to one in the non-negative integers. Since $x_1$ is an integer, $x_1 > 1 \implies x_1 \geq 2$. Similarly, since $x_2$ and $x_3$ are integers, $x_2 > 1 \implies x_2 \geq 2$ and $x_3 > 3 \implies x_3 \geq 4$. Let
\begin{align*}
y_1 & = x_1 - 2\\
y_2 & = x_2 - 2\\
y_3 & = x_3 - 4\\
y_4 & = x_4
\end{align*}
Then each $y_k$, $1 \leq k \leq 4$, is a non-negative integer. Substituting $y_1 + 2$ for $x_1$, $y_2 + 2$ for $x_2$, $y_3 + 4$ for $x_3$, and $y_4$ for $x_4$ in equation 1 yields
\begin{align*}
y_1 + 2 + y_2 + 2 + y_3 + 4 + y_4 & = 12\\
y_1 + y_2 + y_3 + y_4 & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the non-negative integers with
$$\binom{4 + 3}{3} = \binom{7}{3}$$
solutions.
Best Answer
Here is a different way to break it down $$ x_1\in\{1,2,3,4,5\} $$ and given $x_1$ we then have $x_1+x_2<15$ and $x_2>6$ combined as $$ 6<x_2<15-x_1 $$ And whenever $x_1$ and $x_2$ are given, the value of $x_3$ follows from them.
For $x_1=5$ we then have $x_2\in\{7,8,9\}$ so three choices for $x_2$. Each time $x_1$ is decreased by $1$ we gain one option for $x_2$. Thus we have a total of $$ 3+4+5+6+7 = 25 $$ sets of integer solutions under the given constraints.
I ran the following code snippet in Python which confirmed the figure of 25:
I understand that I did not answer the question using the method required, but I wonder why I find the number of solutions to be $25$ whereas the OP and the other answer find it to be $49$. Did I misunderstand the question in the first place?