We have planes 3x + 5y + 7z = 3 ... (1)
And 9x + 15y + 21z = 9 ⇒ 3x + 5y + 7z = 3. ... (2)
We know that distance between two parallel planes Ax + By + Cz = D1 and Ax + By + Cz = D2 is given by \(d=\Big|\cfrac{D_2-D_1}{\sqrt{A^2+B^2+C^2}}\Big|\).
Since, plane (1) and (2) both are parallel to line 3x + 5y + 7z = 0 .
By comparing plane (1) and (2) with planes Ax + By + Cz = D1 and Ax + By + Cz = D2, we get A = 3, B= 5, C = 7,D1 = 3 and D2 = 3.
Therefore, the distance between both planes is
= 0 unit.
