Question 1009556
Let's form two vectors using the three points.
From the first and second,
(2-(-1),1-0,8-(-4))=(3,1,12)
From the first and third,
(-1-(-1),1-0,-1-(-4))=(0,1,3)
Now find the cross product of these two vectors, that will be a vector normal to the plane containing the three points.
N=(3-12)i+(0-9)j+(3-0)k
N=(-9)i+(-9)j+(3)k
N=-3i-3j+k
So then using any of the points (I use the first),
{{{-3(x+1)-3(y-0)+1(z+4)=0}}}
{{{-3x-3-3y+z+4=0}}}
{{{-3x-3y+z=-1}}}
{{{3x+3y-z=1}}}
or
{{{z=3x+3y-1}}}