Find three real orthogonal matrices of order $3$ having all integer entries.

Let $A$ be such a matrix, and $a_1$, $a_2$ and $a_3$ its columns. Given that $A$ is orthogonal means that $$\langle a_i,a_j\rangle =\left\{\begin{array}{ll}1&\text{ if } i=j\\0&\text{ if }i\neq j\end{array}\right.$$ So in particular $\langle a_i,a_i\rangle=1$. Because the $a_i$ have all integer coefficients, it follows that $$a_1,a_2,a_3\in\{(1,0,0),(-1,0,0),(0,1,0),(0,-1,0),(0,0,1),(0,0,-1)\},$$ leaving six options for each of $a_1$, $a_2$ and $a_3$. For any given value of $a_1$ there are precisely $4$ values of $a_2$ such that $\langle a_1,a_2\rangle=0$. For any given values of $a_1$ and $a_2$ there are precisely $2$ values of $a_3$ such that $\langle a_1,a_3\rangle=\langle a_2,a_3\rangle=0$. Hence there are $6\times4\times2=48$ such matrices in total.