Let the rotation axis be a line through a fixed point $(x_0,y_0,z_0,1)^T$ and an infinite point (direction of the line) $(a,b,c,0)^T$, without loss of generality, we assume $a^2+b^2+c^2=1$. Use right-handed rule for rotation, then a general homogeneous rotation with angle $\theta$ can be obtained as:
$$
\boldsymbol{R}^{3D}\left(x_0,y_0,z_0,a,b,c,\theta\right)=\mathscr{C}_1+\left(\sin\theta\cdot\mathscr{A}_2- \left(1-\cos\theta\right)\cdot\mathscr{O}_3\right)\cdot \mathscr{T}_4
$$
where:
$$\mathscr{C}_1=
\left[\begin{array}{*{20}{c}}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&2-\cos\theta
\end{array}\right]$$
$$\mathscr{A}_2 = {\begin{array}{c}
\underbrace{
\begin{array}{c}
{{\left[ {\begin{array}{*{20}{c}}
{\color{blue}0}&{\color{blue}-c}&{\color{blue}b}&0\\
{\color{blue}c}&{\color{blue}0}&{\color{blue}-a}&0\\
{\color{blue}-b}&{\color{blue}a}&{\color{blue}0}&0\\
0&0&0&0
\end{array}} \right]}}
\end{array}
} \\ \text{Antisymmetric matrix}
\end{array}}
$$
$$
\quad\mathscr{O}_3= \begin{array}{c}
\underbrace{ {I - {\left[ {\begin{array}{*{20}{c}}
a\\
b\\
c\\
0
\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}
a&b&c&0
\end{array}} \right]}} }\\
\text{Orthographic parallel projection} \end{array} $$
$$ \mathscr{T}_4=\begin{array}{c}
\underbrace{\left[ {\begin{array}{*{20}{c}}
1&0&0&{ - {x_0}}\\
0&1&0&{ - {y_0}}\\
0&0&1&{ - {z_0}}\\
0&0&0&1
\end{array}} \right]}\\ \text{Translation}\end{array}$$
The L-C formulation of homogeneous 3D rotation is similar to Rodrigues' but they are not the same in essence. Details are available here:
A submission on homogeneous rotation to arXiv.org
For your case, substitute $x_0=y_0=z_0=0$, $a=b=0$, $c=1$ and $\theta=\dfrac{\pi}{2}$ into the L-C formulation then you can obtain the desired homogeneous rotation matrix.