= Solution
{c}
If $T$ is an <orthogonal transformation> and $e_1,e_2,e_3$ is an <orthonormal basis>, then
$$
\langle Te_i,Te_j\rangle=\langle e_i,e_j\rangle=\delta_{ij}.
$$
The images form an <orthonormal set>, hence are <linearly independent> and form an <orthonormal basis> in <dimension> three.
Conversely, suppose the images $Te_1,Te_2,Te_3$ of one chosen <orthonormal basis> form an <orthonormal basis>. Write $u=\sum_i a_ie_i$ and $v=\sum_j b_je_j$. By <linearity> of $T$ and <bilinearity>,
$$
\langle Tu,Tv\rangle=\sum_{i,j}a_ib_j\langle Te_i,Te_j\rangle
=\sum_i a_ib_i=\langle u,v\rangle.
$$
Thus \b[mapping one <orthonormal basis> to an <orthonormal basis> is equivalent to preserving every <inner product>]. In the converse, a single chosen <orthonormal basis> suffices.
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