<s>[INST] 3D Perspective projection I have this following question to answer, however I am not sure how I should combine my calculation into one final answer.

Suppose the Centre of Projection in a viewing space is at an offset
  $(0, 0, -5)$ from $(0,0,0)$, and the view plane is the $UV$ plane containing
  $c$. Find the transform matrix for the perspective projection, and give
  the projected Word Coordinate point $(10,-20,-10)$ on the view plane.

So this is the transformation matrix: with $1/d=0.2$. ($1/5$)
$$
\begin{pmatrix}
1 & 0 & 0 &  0 \\
0 & 1 & 0 &  0 \\
0 & 0 & 0 &  0 \\
0 & 0 & 1/d & 1
\end{pmatrix}
$$
Because the view plane is placed at $z=0$ we use the following similar triangles: 
$$
x' = x/(z/d+1), \quad 
y' = y/(z/d+1), \quad
z'=z/(z/d+1), \quad
w=z/d
$$
My question is, what values do I use for the $x$,$y$,$z$. If using the project world coordinates $x'$, $y'$, $z'$ will be undefined due to the $0$. Once I have the values for $x'$, $y'$, $z'$ do I multiple them by the transformation matrix?
Thanks for any help 
 [/INST] Let me try based on the theories described in the arXiv.org article Unified Framework of Elementary Geometric Transformation Representation.
A so-called perspective projection in this specific problem can be realized by a central projection defined in definition 3.2 (pages 8-9) and formulated in equation (3.1) or (see definitions and formulations in line No. 1 of Table 1 in page 9 of  the arXiv.org article Unified Framework of Elementary Geometric Transformation Representation.
As we know from the problem, the projection center $C=(0,0,-5)$ and view plane is $xOy$ axis plane of the 3D world coordinate system. So for the central projection in homogenous matrix form to be found, the homogenous coordinate of its projection center $(s)=(0,0,-5,1)^T$ in column vector convention, and that of its interface plane $z=0$ which is written into $(\pi)=(0,0,1,0)^T$, are all readily available. So the homogeneous matrix of the desired central projection per definition 3.2 (pages 8-9) and   equation (3.1) (or definition and formulation in line No. 1 of Table 1 in page 9) of  the arXiv.org article Unified Framework of Elementary Geometric Transformation Representation can be obtained easily:
$$T=\left[
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right]
-\dfrac{\left[
\begin{array}{c}
 0 \\
 0 \\
 -5 \\
 1 \\
\end{array}
\right]\cdot \left[
\begin{array}{cccc}
 0 & 0 & 1 & 0 \\
\end{array}
\right]}{\left[
\begin{array}{cccc}
 0 & 0 & -5 & 1 \\
\end{array}
\right]\cdot \left[
\begin{array}{c}
 0 \\
 0 \\
 1 \\
 0 \\
\end{array}
\right]}=\left[
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & \dfrac{1}{5} & 1 \\
\end{array}
\right]$$
It is easy to apply the obtained central projection to the given $(10,-20,-10)$ of which a column vector homogenous coordinate is :
$$X= \left[
\begin{array}{c}
 10 \\
 -20 \\
 -10 \\
 1 \\
\end{array}
\right]$$
then 
$$X'=\left[
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & \dfrac{1}{5} & 1 \\
\end{array}
\right]\cdot \left[
\begin{array}{c}
 10 \\
 -20 \\
 -10 \\
 1 \\
\end{array}
\right]=\left[
\begin{array}{c}
 10 \\
 -20 \\
 0 \\
 \color{red}{-1} \\
\end{array}
\right]\approx \color{red}{({\text{nonzeor scalar } -1})} \left[
\begin{array}{c}
 -10 \\
 20 \\
 0 \\
 \color{red}1 \\
\end{array}
\right]$$
So the final projected point has the 3D Euclidean coordinate of $\color{red}{(-10,20,0)}$. 
Similarly for your point $(10,-10,10)$ you will obtain 
$$\left[
\begin{array}{c}
 10 \\
 -10 \\
 0 \\
 \color{red}3 \\
\end{array}
\right]\approx \color{red}{({\text{nonzeor scalar } 3})} \left[
\begin{array}{c}
 \dfrac{10}{3} \\
 -\dfrac{10}{3} \\
 0 \\
 \color{red}1 \\
\end{array}
\right]$$
the projected 3D Euclidean coordinate of which should be $\left(\dfrac{10}{3}, -\dfrac{10}{3},
 0\right)$.
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