Furthering the equation i get

$\displaystyle 12(x-x0) + 21(y-y0) -22(z-z0) = 0 $

where as shown i obtained 12, 21 and -22 by calculating the normal. I then substitute a vector in the plane, i.e 4,1,6), giving you an equation

$\displaystyle 12(x-4) + 21(y-1) -22(z-6) = 0. $

The equation can now be rewritten to the form

$\displaystyle ax+by+cz = d $

then

$\displaystyle 12x+21y+-22z = d $

I'm not sure how to reorganize that equation to evaluate the d mentioned above. Any help /push in the right direction would be great!

Once this is done then I should be able to check my equation by seeing whether each of the given three points in the plane satisfies it right ?

Thanks a bunch