Does it look like the correct 3D phase portrait? What I am talking about: https://en.m.wikipedia.org/wiki/Phase_space
You can test it by yourself in my site(I have made it on three.js),you can move it and scale
https://imaimachi.sumy.ua/levels

I have a system of kinetic equations:
$$\dfrac{dx}{dt}= 2x+32-z $$
$$\dfrac{dy}{dt}= 5y+x+17 $$
$$\dfrac{dz}{dt}= 6x-2z+19y $$
And I had an idea to build a phase portrait of it. I made it using isoclines method. I found isocline equations for each angle from 1 to 360 degrees. I drawn it using three js in browser. 
What did I get:

I have never done a 3D phase portraits and actually never worked with 3D graphics. 
So, does it look like the correct phase portrait for this system?
Update. What are the steps literally:
I make a system
$$\dfrac{dz}{dx}= \dfrac{6x-2z+19y}{2x+32-z} $$
$$\dfrac{dz}{dy}= \dfrac{6x-2z+19y}{5y+x+17} $$
Equals constant for each angle:
$$\dfrac{6x-2z+19y}{2x+32-z} = tg \alpha $$
$$\dfrac{6x-2z+19y}{5y+x+17} = tg \alpha $$
For each $\alpha$ from 1 to 360 I get solution for this system in form
$$y=a(b*x+c)$$
$$z=d(e*x+h)$$
And then draw the isocline parametric equation $(x,\space a(b*x+c),\space d(e*x+h))$
Update 2. The second test
I have drawn a phase portrait for this system
$$\dfrac{dx}{dt}= 5y+3 $$
$$\dfrac{dy}{dt}= x+16y-z-9 $$
$$\dfrac{dz}{dt}= 12x-18y+z $$
Phase portrait 

At this time it looks like of the type of equilibrium point 

But the problem is that the system has only one equilibrium point $x=\dfrac{3}{5},\space y=-\dfrac{3}{5},\space z=-18$. In the graph I made it seems like (0,0,0)
 A: A general methodology for such systems :
Write your differential system under the form:
$$V'=AV+B \tag 1 $$
for an appropriate matrix $A$ and vector $B$.
The RHS is affine. Transform it into a linear system :
$$W'=AW \ \ \text{where} \ \ W:=V-V_0$$
by substracting to (1) equation :
$$0=AV_0+B$$
[Such a constant vector $V_0$ exists due to the invertibility of $A$] .
then express $W$ in the form of the exponential of a matrix applied to a vector :
$$W=\exp(tA) W_{ini}$$
where $W_{ini}$ is obtained through initial conditions, finally giving :

$$V=\exp(tA)(V_{ini}-V_0)+V_0$$

(of course : $V_{ini}$ are initial conditions on $V$).
Numerical application : 
$$A=\begin{pmatrix}2 &0 &-1\\1& 5 &0\\6&19& -2\end{pmatrix}, \ \ B=\begin{pmatrix}32\\17\\0\end{pmatrix} , \ \ V_0=\begin{pmatrix}-643/9\\98/9\\-998/9\end{pmatrix}  $$
then use a Computer Algebra System for obtaining $\exp(tA)$ using eigenvalues $    3.8920, 2.1725, -1.0644$ of $A$.
A: Yes, it is correct.
The first one is exception - it’s wrong, because I rotated each part of integral curve due to z axis, instead of x, and some logics in my code was wrong. 
The second one is perfectly correct, just look:

X - RED
Y - GREEN
Z - BLUE
Look at the x equation- it depends only on y, and the dependence is linear. Now look on the “2D” case of x/y to see better:

I think it’s obvious that all is correct.
So, now I can see how the system behaves, haven’t solved it directly - this is the phase portrait.

This is how z depends on x

