# 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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To find the equation of the plane we need to find its normal vector.
Let us call the given points as: P(2,1,2), Q(3,-8,6), R(-2,-3,1)
Note that $\stackrel{\to }{PQ}$ and $\stackrel{\to }{PR}$ are vectors in the plane
Hence, their cross product will give us normal vector to the plane
$\stackrel{\to }{PQ}=\left(3,-8,6\right)-\left(2,1,2\right)=\left(3-2,-8-1,6-2\right)=\left(1,-9,4\right)$
$\stackrel{\to }{PR}=\left(-2,-3,1\right)-\left(2,1,2\right)=\left(-2-2,-3-1,1-2\right)=\left(-4,-4,-1\right)$
$n=\stackrel{\to }{PQ}×\stackrel{\to }{PR}=|\begin{array}{ccc}i& j& k\\ 1& -9& 4\\ -4& -4& -1\end{array}|=\left(25,-15,-40\right)$
Equation of a plane passing through the point (a,b,c) and having normal vector (l,m,n) is
$l\left(x-a\right)+m\left(y-b\right)+n\left(z-c\right)=0$
We found the normal vector in the previous cell. For a point on the plane, we can choose any of the three given points, I will choose P(2,1,2)
Equation of the plane is
$25\left(x-2\right)-15\cdot \left(y-1\right)-40\cdot \left(z-2\right)\right]=0$
Divide throughout by 5
$5\left(x-2\right)-3\cdot \left(y-1\right)-8\cdot \left(z-2\right)=0$
$5x-3y-8z+9=0$
Result:
$5x-3y-8z+9=0$