= Solution
{c}
A <Givens rotation> acts as the identity except on coordinates $i,j$, where its block can be written
$$
\begin{pmatrix}\cos\theta&\sin\theta\\
-\sin\theta&\cos\theta\end{pmatrix}.
$$
The block has orthonormal columns because $\cos^2\theta+\sin^2\theta=1$ and their inner product is zero, so $J^TJ=I$. For $r=\sqrt{v_i^2+v_j^2}>0$, choose $\cos\theta=v_i/r$, $\sin\theta=v_j/r$. The new coordinates are $(r,0)$, eliminating the $j$th entry. If both entries are zero, the identity already works.
For the specified vector, one suitable matrix is
$$
\boxed{Q=\begin{pmatrix}
4/5&0&-3/5\\3/5&0&4/5\\0&-1&0
\end{pmatrix},\qquad Qv=\begin{pmatrix}0\\5\\0\end{pmatrix}.}
$$
Its rows are orthonormal, so it is orthogonal; in fact its <determinant> is one. The positive nonzero output component must be five because an <orthogonal matrix> preserves <Euclidean norm>.
\b[The matrix is not unique.] Left multiplication by any <orthogonal matrix> fixing the second coordinate vector preserves this output. For example arbitrary rotations of the first and third coordinates provide infinitely many alternatives.
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