= Solution
Let $Q\in\mathrm{SO}(3,\mathbb R)$. Since $Q^TQ=I$ and $\det Q=1$,
$$
\det(Q-I)=\det Q\,\det(I-Q^T)=\det(I-Q)=-\det(Q-I),
$$
so $\det(Q-I)=0$. Choose a <unit vector> $e$ with $Qe=e$. Its <orthogonal complement> is an <invariant subspace>: for $v\perp e$, $e\cdot Qv=(Q^Te)\cdot v=e\cdot v=0$. In an <orthonormal basis> $(e,e_1,e_2)$, $Q$ therefore has a scalar block $1$ and a two-dimensional <orthogonal matrix> block of <determinant> one. Such a block has columns $(\cos\theta,\sin\theta)^T$ and $(-\sin\theta,\cos\theta)^T$. Consequently $Q$ is a <rotation in three dimensions> about the axis $\mathbb Re$ through angle $\theta$, including the identity when $\theta=0$.
To construct the <Householder reflections>, let $S_0$ reflect in the plane spanned by $e,e_1$ and let $S_\psi$ reflect in the plane spanned by $e,\cos\psi\,e_1+\sin\psi\,e_2$. They fix $e$, while on $e^\perp$ their <matrices> are
$$
S_0=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad S_\psi=\begin{pmatrix}\cos2\psi&\sin2\psi\\\sin2\psi&-\cos2\psi\end{pmatrix}.
$$
Their product is the planar <rotation matrix> through $2\psi$. Taking $\psi=\theta/2$ proves $\boxed{Q=S_{\theta/2}S_0}$. This gives the <composition of two plane reflections> explicitly; each is a <Householder reflection> $I-2nn^T$ for the corresponding unit plane normal $n$.
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