Can you solve this 5 radicals problem?

$\sqrt{x - y - 3} + \sqrt{y ^ {2} - 9y + 3x} =$ $\sqrt[3]{30x - 30 y + 35} + \sqrt{3 + y - x}$

From the square root restrictions we have:

x - y - 3 >= 0 (#1)

$y ^ {2} - 9y + 3x >= 0$ (#2)

3 + y - x >= 0 (#3)

From (1): x - y >= 3 (#4)

From (3): x - y <= 3 (#5)

From (4) and (5) : x - y = 3 i.e. y = x - 3 (#6)

Substituting y by (#6) in the original equation we get:

$\sqrt{x - x + 3 - 3} + \sqrt{(x - 3) ^ {2} - 9(x - 3) + 3x} = \sqrt[3] {30x - 30x + 90 + 35} + \sqrt{3 + x - 3 - x}$

$\sqrt{0} + \sqrt{(x ^ {2} - 6x + 9 - 9x + 27 + 3x)} = \sqrt[3] {125} + \sqrt{0}$

$\sqrt{x ^ {2} - 12x + 36} = 5$

|x - 6| = 5

x - 6 = 5 or x - 6 = -5

if x - 6 = 5, then x = 11, y = 11 - 3 = 8, z = $\sqrt[5] {2.4(11) + 0.7(8)} = \sqrt[5] {32} = 2$

if x - 6 = -5, then x = 1, y = 1 - 3 = -2, z = $\sqrt[5] {2.4(1) + 0.7(-2)} = \sqrt[5] {1} = 1$

Total of all values of z = 1 + 2 = 3

$Answer = \boxed{3}$