Find an equation of a plane which contains the points: $(0,0,3),(3,2,1)0$, and $(6,2,0)$
I know I need a vector in order to use the equation $d=ax_0$+b$y_0$+c$z_0$
Now, could I just select any two points as my terminal and initial, say, (0,0,3) and (3,2,1) subtract the two and create a "n" vector, and then just go from there? Or is there something I must do will all three points?
If the points are $P,Q,R$ you need to form vectors $\vec{v_1}=\overrightarrow{PQ}$ and $\vec{v_2}=\overrightarrow{PR}$ and then compute the cross product and thus the normal vector $$ \vec n=\vec{v_1}\times\vec{v_2} $$ The the equation you seek will emerge from expanding the dot product $$ \vec n\cdot \left( \begin{pmatrix} x\\y\\z \end{pmatrix}-\overrightarrow{OP}\right)=0 $$ where $\overrightarrow{OP}$ is the vector sharing the same coordinates as the point $P$, essentially pointing from the origin of the coordinate system to the point $P$.