Step 1: Understanding the Topic
This question deals with a system of linear equations. For a system to have infinitely many solutions, the equations must be linearly dependent. This means one of the equations can be expressed as a linear combination of the others. Here, the third equation must be a linear combination of the first two.
Step 2: Key Approach - Linear Combination
We will set the third equation to be a weighted sum of the first two equations, using constants $\lambda$ and $\mu$. By comparing the coefficients of $x$, $y$, $z$, and the constant terms, we can solve for these constants and subsequently find the values of $a$ and $b$.
Step 3: Detailed Calculation
Let the third equation be represented as $\lambda \times (\text{Eq. 1}) + \mu \times (\text{Eq. 2})$:
\[
\lambda(x + 2y + 2z) + \mu(2x + 3y + 2z) = \lambda(1) + \mu(2)
\]
\[
(\lambda + 2\mu)x + (2\lambda + 3\mu)y + (2\lambda + 2\mu)z = \lambda + 2\mu
\]
Now, we compare this with the given third equation: $ax + 5y + bz = b$.
Coefficient of $x$: $a = \lambda + 2\mu$
Coefficient of $y$: $5 = 2\lambda + 3\mu$
Coefficient of $z$: $b = 2\lambda + 2\mu$
Constant term: $b = \lambda + 2\mu$
From equations (1) and (4), we see that $a=b$.
From equations (3) and (4), we have:
\[
2\lambda + 2\mu = \lambda + 2\mu \implies \lambda = 0
\]
Now that we have $\lambda = 0$, we can substitute it into equation (2) to find $\mu$:
\[
5 = 2(0) + 3\mu \implies 3\mu = 5 \implies \mu = \frac{5}{3}
\]
With $\lambda$ and $\mu$ found, we can find $a$ and $b$:
\[
a = \lambda + 2\mu = 0 + 2\left(\frac{5}{3}\right) = \frac{10}{3}
\]
\[
b = \lambda + 2\mu = 0 + 2\left(\frac{5}{3}\right) = \frac{10}{3}
\]
Finally, we need to find the value of $a+b$.
\[
a + b = \frac{10}{3} + \frac{10}{3} = \frac{20}{3}
\]
Step 4: Final Answer
For the system to have infinite solutions, $a = 10/3$ and $b = 10/3$. Their sum is:
\[
\boxed{a + b = \frac{20}{3}}
\]