To solve this system of equations, we can use the method of substitution or elimination. Let's start by solving the first two equations for x and y in terms of z:

2x + 4y + 4z = 30
x + 3y + 2z = 150

First, solve the second equation for x in terms of y and z:
x = 150 - 3y - 2z

Now substitute x into the first equation:
2(150 - 3y - 2z) + 4y + 4z = 30
300 - 6y - 4z + 4y + 4z = 30
-2y = -270
y = 135

Now that we have found the value of y, we can substitute it back into the second equation to find the value of x:
x + 3(135) + 2z = 150
x + 405 + 2z = 150
x = -255 - 2z

Now substitute the values of x and y into the third equation and solve for z:
2(-255 - 2z) + 10(135) + 9z = 110
-510 - 4z + 1350 + 9z = 110
5z = -750
z = -150

Now that we have found the value of z, we can substitute it back into the equations to find the values of x and y:
x = -255 - 2(-150) = -255 + 300 = 45
y = 135

Therefore, the solution to the system of equations is x = 45, y = 135, and z = -150.