1. Equation of the plane is
⇒ (x – 2)(-21) – (y – 1)(7 + 2) + z(0 + 3) = 0
⇒ 21x + 42 – 9y + 9 + 3z = 0 ⇒ -21x – 9y + 3z + 51 = 0
⇒ 7x + 3y – z = 17.
2. Vector form is \(\bar r\).(7i + 3j – k) = 17 _____(1)
3. Now, |7i + 3j – k| = \(\sqrt{49+9+1} = \sqrt{59}\)
Dividing equation (1) by \(\sqrt{59}\), we get
Therefore the above equation is the normal form of the plane. Then \(\frac{7i+3j-k}{\sqrt{59}}\)is the unit vector perpendicular to the plane and \(\frac{17}{\sqrt{59}}\) is the perpendicular distance from the origin.