(i) The equation of the plane is
\(\frac{x}{2}+\frac{y}{3}+\frac{z}{4} = 1\)
(ii) The equation of the plane is 2x – 3y + 4z = 4
Hence the distance = \(\frac{2(-1)-3(-2)+4(3)-4}{\sqrt{4+9+16}}\)
\( = \frac{-2+6+12-4}{\sqrt{4+9+16}}\)
= \(\frac{12}{\sqrt{29}}\)