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:
n = 0
for x1 in range(1,16):
for x2 in range(1,16):
for x3 in range(1,16):
if x1 < 6 and x2 > 6 and x1+x2+x3 == 15:
n += 1
print n, ":", x1, x2, x3
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?
How many integers solutions of $$x_1 + x_2 + x_3 + x_4 = 28$$ are there that satisfy $-10 \leq x_k \leq 20$ for $1 \leq k \leq 4$?
We can convert this to an equivalent problem in the nonnegative integers.
$$x_1 + x_2 + x_3 + x_4 = 28 \tag{1}$$
Let $y_k = x_k + 10$, where $1 \leq k \leq 4$. Then each $y_k$ is a nonnegative integer satisfying $0 \leq k \leq 30$. Substituting $y_k - 10$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields
\begin{align*}
y_1 - 10 + y_2 - 10 + y_3 - 10 + y_4 - 10 & = 28\\
y_1 + y_2 + y_3 + y_4 & = 68 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. A particular solution of equation 2 corresponds to the placement of $3$ addition signs in a row of $68$ ones. The number of solutions of equation 1 is
$$\binom{68 + 3}{3} = \binom{71}{3}$$
since we must choose which $3$ of the $71$ positions required for $68$ ones and $3$ addition signs will be filled with addition signs.
From these, we must subtract those cases that violate one or more of the restrictions that $y_k \leq 30$, $1 \leq k \leq 4$. Notice that at most two of these restrictions can be violated simultaneously since $3 \cdot 31 = 93 > 68$.
Suppose $y_1 > 30$. Then $y_1' = y_1 - 31$ is a nonnegative integer. Substituting $y_1' + 31$ for $y_1$ in equation 2 yields
\begin{align*}
y_1' + 31 + y_2 + y_3 + y_4 & = 68\\
y_1' + y_2 + y_3 + y_4 & = 37 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{37 + 3}{3} = \binom{40}{3}$$
solutions. By symmetry, there are an equal number of solutions for each of the four variables that could violate the restriction $y_k \leq 30$. Hence, there are
$$\binom{4}{1}\binom{40}{3}$$
cases in which one of the restrictions is violated.
However, if we subtract this number from the total, we will have subtracted too much since we will have subtracted each case in which two of the variables violate the restrictions twice, once for each variable we designated as the variable that violated the restriction. We only want to subtract those cases once, so we must add them back.
Suppose $y_1, y_2 > 30$. Let $y_1' = y_1 - 31$; let $y_2 = y_2 - 31$. Then $y_1'$ and $y_2'$ are nonnegative integers. Substituting $y_1' + 31$ for $y_1$ and $y_2' + 31$ for $y_2$ in equation 2 yields
\begin{align*}
y_1' + 31 + y_2' + 32 + y_3 + y_4 & = 68\\
y_1' + y_2' + y_3 + y_4 & = 6 \tag{4}
\end{align*}
Equation 4 is an equation in the nonnegative integers with
$$\binom{6 + 3}{3} = \binom{9}{3}$$
solutions. By symmetry, there are an equal number of solutions for each of the $\binom{4}{2}$ pairs of variables that could violate the restrictions. Hence, there are
$$\binom{4}{2}\binom{9}{3}$$
cases in which two of the variables violate the restrictions.
Thus, by the Inclusion-Exclusion Principle, the number of solutions that satisfy the restrictions is
$$\binom{71}{3} - \binom{4}{1}\binom{40}{3} + \binom{4}{2}\binom{9}{3}$$
Best Answer
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, $$ M = {17\choose 2} - \left( 2{11\choose 2} + {9\choose2}\right) + \left({5\choose 2} + 2{3\choose 2}\right) = 6. $$