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**terrorsquid** I have three points $\displaystyle A = (2,0,1)$, $\displaystyle B = (1,1,2)$ and $\displaystyle C = (0,0,3)$ that are the vertices of a triangle in $\displaystyle \mathbb{R}^3$. I am trying to find the equation of the plane on which the triangle lies in parametric and cartesian form.

I have an answer written down:

$\displaystyle (x,y,z) = (2,0,1) + \lambda (-1,1,1) + \mu (-1,-1,1)$

But I'm not sure what method was used to achieve this answer. The way I did it is:

$\displaystyle \vec{AB} = (-1,1,1)$

$\displaystyle \vec{BC} = (-1,-1,1)$

$\displaystyle \vec{AB} \times \vec{BC} = 2x-0y+2z$

$\displaystyle \therefore (x,y,z) = 2(x-2)+2(z-1)$

$\displaystyle x+z = 3$ (cartesian)

Is this right and how do I get the parametric form?