1. Given, \(\bar r\).(2i – 3j + 5k) = 7 and if we substitute \(\bar r\) = xi + yj + zk Then we get the Cartesian equation as 2x – 3y + 5z – 7 = 0.
2. The equation of a plane parallel to the above plane differ only by a constant, therefore let the equation be 2x – 3y + 5z + k = 0.
⇒ 6 – 12 – 5 + k = 0 ⇒ k = 11
Therefore the equation is 2x – 3y + 5z + 11 = 0
3. The distance between the parallel planes