= Solution
At the angles in part (f),
$$
R_1=\begin{pmatrix}0&-1&0\\0&0&1\\-1&0&0\end{pmatrix}.
$$
Let the two <diagonal matrices> be $D=\operatorname{diag}(d_1,d_2,d_3)$ and $D'=\operatorname{diag}(e_1,e_2,e_3)$. Multiplication gives
$$
D'R_1D=\begin{pmatrix}0&-e_1d_2&0\\0&0&e_2d_3\\-e_3d_1&0&0\end{pmatrix}.
$$
The vector condition is exactly $ce_2d_3=b$. For $c\ne0$, the general form is therefore
$$
\boxed{\widetilde R=\begin{pmatrix}0&p&0\\0&0&b/c\\q&0&0\end{pmatrix},\qquad p,q\in\mathbb R.}
$$
Every pair $p,q$ is attainable, for example with $D=I$ and $D'=\operatorname{diag}(-p,b/c,-q)$. The transformation need not be orthogonal or invertible; the question asks only for diagonal rescalings with this one vector image. If $c=0$ and $b\ne0$, no linear map can satisfy the condition. If $c=b=0$, the $(2,3)$ entry is arbitrary as well.
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