# closest point on a plane to another point in $\mathbb{R}^3$

Given $4$ points in $\mathbb{R}^3$: $A(0,2,4);B(-2,6,-2);C(2,-4,8);D(10,2,0)$, find the line equation $AK$ when $K$ is the projection of $D$ on the plane $ABC$.

The first thing I did was find the equation for the plane made up of the points $A$,$B$, and $C$. I found this to be: $5x +y -z +18$ after choosing an arbitrary point $M$ and setting $(AM \times AB)⋅AC = 0$.

next I figured the line normal to the plane I just found going through point $D$ would be: $(10,2,0) + (5,1,-1)t$ or in parametric form:

$x=10+5t$

$y=2+t$

$z=-t$

I plugged these values into my plane equation $5x +y -z=-18$ to find the point of intersection and found that for $t=\frac{-70}{27}$ I get the point K $(\frac{-80}{27},\frac{-16}{27},\frac{-70}{27})$.

Then I thought to present this line in the vector format, $A + (K-A)t$ but this isn't anywhere close to the answer:

$(0,2t,2+2t)$

What am I doing wrong?

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It seems that your plane equations is not correct. I got the following answer: $$5x+y-z+2=0$$