I follow the approach here, which is the answer by user84413, which I explain in more detail.
As seen before, using $y_1=x_1$, $y_2=x_2-4$, $y_3=x_3-7$ we may equivalently consider $y_1+y_2+y_3=7$ subject to $0\le y_1\le 6$, $0\le y_2\le 5$ and $0\le y_3\le 7$.
Hence, we want to know the number of elements in $\mathcal S=(\mathcal T_{7,1,7}\cup\mathcal T_{7,2,6}\cup\mathcal T_{7,3,8})^{\mathrm c}\subseteq\mathcal T_7$, where $\mathcal T_s=\{y\in\mathbb Z_{\ge 0}^3:\sum_{i=1}^3y_i=s\}$ and $\mathcal T_{s,i,u}=\{y\in\mathcal T_s:y_i\ge u\}$. This way to define $\mathcal S$ using de Morgan's law allows to use the Inclusion-Exclusion Principle. To be precise, let $\mathcal S_1=\mathcal T_{7,1,7}$, $\mathcal S_2=\mathcal T_{7,2,6}$ and $\mathcal S_3=\mathcal T_{7,3,8}$. Then we have
$$|\mathcal S|=|\mathcal T_7|-\sum_{k=1}^{3}(-1)^{k+1}\sum_{1\le i_1<\cdots<i_k\le 3}\left|\bigcap_{j=1}^{k}\mathcal S_{i_j}\right|.$$
Notice that $|\mathcal T_s|=\binom{s+3-1}{s}=\binom{s+2}{s}$ using starts and bars. Further, notice that $$\mathcal T_{s,1,u}=\left\{y\in\mathbb Z_{\ge 0}^3:\sum_iy_i=s,y_1\ge u\right\}
=\{(z_1+u,z_2,z_3):z\in\mathcal T_{s-u}\},$$
which yields $|\mathcal T_{s,i,u}|=|\mathcal T_{s-u}|$ since we chose $i=1$ only for brevity. Analogously, we get $|\mathcal T_{s,i,u}\cap\mathcal T_{s,j,v}|=|\mathcal T_{s-u-v}|$ for $i\neq j$ and $|\mathcal T_{s,1,u}\cap\mathcal T_{s,2,v}\cap\mathcal T_{s,3,w}|=|\mathcal T_{s-u-v-w}|$. Notice that $\mathcal T_{s}=0$ for $s<0$ and hence $\mathcal S=\binom{7+2}{7}-\binom{0+2}{0}-\binom{1+2}{1}=\binom{9}{2}-1-3=32$.