Question

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 units is:

Answer 1

Equation of Plane $ABC$ will be

$|x-1y-1z-1124-2-11|=0$

Equation will be

$6(x-1)-(y-1)6+(z-1)(+3)=0$

$2x-2y+z=1$

$2x-2y+z-1=0$

So, Plane equation parallel to $ABC$ will be $2x-2y-zx+k=0$

distance b/w two planes $=d=\left|\frac{k+1}{\sqrt{2^2+{(-2)}^2+{\left(1\right)}^2}}\right|=2$

So, $\left|\frac{k+1}{3}\right|=2$

$|k+1|=6$

$k=5$ or $k=-7$

So, equation of plane will be either

$2x-2y+z+5=0$

or $2x-2y+z-7=0$