Answer
Not on the same line
Work Step by Step
Step 1:
Theorem:
The line in $\mathbb{R}^3$ that passes through the point $\mathbf{P_0}=(x_0,y_0,z_0)$ and is parallel to the nonzero vector $\mathbf{v}=\langle a,b,c \rangle = ai+bj+ck$ has parametric equations:
$$\begin{aligned} x&=x_0+at\\ y&=y_0+bt\\ z&=z_0+ct \end{aligned}$$
The vector equations of these lines can be written as:
$$\langle a,b,c \rangle = \langle x_0+at, y_0+bt, z_0+ct \rangle$$
or
$$\langle a,b,c \rangle = \langle x_0,y_0,z_0 \rangle + t\langle a,b,c \rangle$$
Step 2:
We notice that if the line passes through $\mathbf{P_1}=(x_1,y_1,z_1)$, then it is parallel to $\mathbf{v}=\overrightarrow{\mathbf{P_0P_1}}$ and the equation of the line can be written as:
$$\begin{aligned} \langle a,b,c \rangle &= \langle x_0,y_0,z_0 \rangle + t\langle x_1-x_0, y_1-y_0, z_1-z_0 \rangle\\ &= \langle x_0,y_0,z_0 \rangle + t\overrightarrow{\mathbf{P_0P_1}} \end{aligned}$$
For $\mathbf{P_2}=(6,9,0),\ \mathbf{P_2'}=(9,2,0)$ we have that:
$$\begin{aligned} \overrightarrow{\mathbf{P_2P_2'}} &= \langle 9,2,0 \rangle - \langle 6,9,7 \rangle\\ &= \langle 3,-7,-7 \rangle\\ \mathbf{v} &= \langle 3,-7,-7 \rangle \end{aligned}$$
and the vector equation of the line becomes:
$$\begin{aligned} \langle x,y,z \rangle &= \langle 6,9,0 \rangle + t\langle 3,-7,-7 \rangle\\ &= \langle 6+3t, 9-7t, -7t \rangle \end{aligned}$$
If we evaluate the third point $\mathbf{P_3}=(0,-5,-3)$ we obtain:
$$\begin{aligned} \langle 0,-5,-3 \rangle &= \langle 6+3t, 9-7t, -7t \rangle\\ \Rightarrow \ \ &6+3t=0,\ 9-7t=-5,\ -7t=-3\\ \Rightarrow \ \ &t=-2,\ t=2,\ t=3/7 \end{aligned}$$
Since we obtained different values for the parameter, this point is not on the line that passes through $\mathbf{P_1}$ and $\mathbf{P_2}$.
