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## Main Question or Discussion Point

How can I find the parametric vector form of a cartesian equation under a

Cartestian equation: $$-2x-y+z=6$$

I know to find the parametric vector form we can find any 3 points P, Q and R which satisfy the cartesian equation.

$$ \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix} =P \begin{pmatrix} x_2\\ y_2\\ z_2 \end{pmatrix} =Q \begin{pmatrix} x_3\\ y_3\\ z_3 \end{pmatrix} =R $$

The parametric equation thus becomes:

$$x=P+\lambda(Q-P)+\mu(R-P)$$

This part is clear, but my question is how can we find the points P, Q and R if Q-P has been specified?

$$ \begin{pmatrix} P \end{pmatrix} + \lambda \begin{pmatrix} 1\\ -2\\ 0\\ \end{pmatrix} + \mu \begin{pmatrix} R-P \end{pmatrix} $$

*specific*condition?Cartestian equation: $$-2x-y+z=6$$

I know to find the parametric vector form we can find any 3 points P, Q and R which satisfy the cartesian equation.

$$ \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix} =P \begin{pmatrix} x_2\\ y_2\\ z_2 \end{pmatrix} =Q \begin{pmatrix} x_3\\ y_3\\ z_3 \end{pmatrix} =R $$

The parametric equation thus becomes:

$$x=P+\lambda(Q-P)+\mu(R-P)$$

This part is clear, but my question is how can we find the points P, Q and R if Q-P has been specified?

$$ \begin{pmatrix} P \end{pmatrix} + \lambda \begin{pmatrix} 1\\ -2\\ 0\\ \end{pmatrix} + \mu \begin{pmatrix} R-P \end{pmatrix} $$