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  4. Let A (1,1, 1), B (2, 3,5) and C (-1,0,2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is

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Q. Let $A (1,1, 1), B (2, 3,5)$ and $C (-1,0,2)$ be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is

Three Dimensional Geometry

Solution:

$A (1, 1, 1), B (2, 3, 5), C (-1 , 0, 2)$ direction ratios of AB are $< 1 ,2 ,4 >.$
Direction ratios of AC are $<-2,-1,1 >.$
Therefore, direction ratios of normal to plane ABC are $< 2 ,-3,1 >$
As a result, equation of the plane $ABC$ is $2x - 3y + z = 0.$
Let the equation of the required plane be $2x - 3y + z = k.$ Then
$\left|\frac{k}{\sqrt{4+9+1}}\right|=2\,\,$or $\,\,k=\pm2\sqrt{14}$
Hence, equation of the required plane is $2x - 3y + z + 2 \sqrt{14} = 0$