Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the plane be $l(x+2)+m(y+2)+n(z-2)=0$. It contains (1, 1, 1) and $(1,-1,2)$, so $3l+3m-n=0$ and $3l+m=0$. Hence $l:m:n=1:-3:-6$. Plane: $x-3y-6z+8=0$, i.e. $\frac{x}{-8}+\frac{y}{8/3}+\frac{z}{4/3}=1$. Lengths of intercepts: $a=8,b=\frac{8}{3},c=\frac{4}{3}$. Then $a=3b,b=2c,a+b+c=12$ and $a+3b+3c=20$. All four are correct.