From the first equation, we can rewrite it as x/3 = y/2. Multiplying both sides by 3, we get x = 3y/2.

Substituting this into the second equation, we have 7(3y/2) = 5z. Simplifying, we get 21y/2 = 5z.

Now let's substitute x = 3y/2 into the third equation: 4(3y/2) - 3y - 2z = -24. Simplifying, we get 6y - 3y - 2z = -24. Combining like terms, we have 3y - 2z = -24.

So we have the system of equations:
21y/2 = 5z
3y - 2z = -24

To solve this system, we can use substitution or elimination method.

Using substitution method:
From the first equation, we can rewrite it as z = (21y/2)/5 = 21y/10.
Substituting this into the second equation, we have 3y - 2(21y/10) = -24.
Simplifying, we get 3y - 42y/10 = -24.
Combining like terms, we have (30y - 42y)/10 = -24.
Simplifying further, we get -12y/10 = -24.
Multiplying both sides by 10, we get -12y = -240.
Dividing both sides by -12, we get y = 20.

Substituting y = 20 into the first equation, we have 21(20)/2 = 5z.
Simplifying, we get 210 = 5z.
Dividing both sides by 5, we get z = 42.

Substituting y = 20 and z = 42 into x = 3y/2, we have x = 3(20)/2 = 30.

So the solution to the system of equations is x = 30, y = 20, and z = 42.