I have been trying to figure out how to go about answering this question however I just cannot seem to wrap my head around it.
"Using similar triangle arguments, give the perspective equations for $x$ and $y$ when the image plane is at $z=0$, the camera position (the focal point) is at $d$, the camera is looking out the negative $z$ axis, and the point is at $(X,Y,Z)$. Once you have the equations, then put them into homogenous matrix that gives $(x,y,0) = M (X,Y,Z).$ "
Perhaps this diagram will make it a bit easier to work out the proportions.
Notice that from similar triangles we obtain the proportions:
\begin{equation} \frac{x}{d}=\frac{X}{d-Z} \end{equation} and \begin{equation} \frac{y}{d}=\frac{Y}{d-Z} \end{equation}
Thus
\begin{equation} x=\frac{d}{d-Z}X \end{equation} and \begin{equation} y=\frac{d}{d-Z}Y \end{equation}
As a matrix equation this may be re-written
\begin{equation} \begin{pmatrix} x\\y\\0 \end{pmatrix} = \begin{pmatrix} \frac{d}{d-Z}&0&0\\ 0&\frac{d}{d-Z}&0\\ 0&0&0 \end{pmatrix} \cdot \begin{pmatrix} X\\Y\\Z \end{pmatrix} \end{equation}