# Find an equation of the plane. The plane through the points (2, 1, 2), (3, −8, 6), and (−2, −3, 1)

Find an equation of the plane. The plane through the points (2, 1, 2), (3, −8, 6), and (−2, −3, 1)
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Let the given points be P(2,1,2),Q(3,-8,6), and R(-2,-3,1)
$\stackrel{\to }{PQ}=⟨3-2,-8-1,6-2⟩=⟨1,-9,4⟩$
$\stackrel{\to }{PR}=⟨-2-2,-3-1,1-2⟩=⟨-4,-4,-1⟩$
$\stackrel{\to }{PQ}×\stackrel{\to }{PR}=|\begin{array}{ccc}i& j& k\\ 1& -9& 4\\ -4& -4& -1\end{array}|=\left(9+16\right)i+\left(-16+1\right)j+\left(-4-36\right)k$
$=25i-15j-40k$
Therefore, the normal vector to the plane is
$\stackrel{\to }{n}=⟨25,-15,-40⟩$
Since, the plane passes through all the three points we can choose any point to find its equation. So, the equation of the plane through the point P(2,1,2) with normal vector $\stackrel{\to }{n}=⟨25,-15,-40⟩$ is
$25\left(x-2\right)-15\left(y-1\right)-40\left(z-2\right)=0$
$⇒25x-50-15y+15-40z+80=0$
$⇒25x-15y-40z+45=0$