Step 1: Know the meaning of infinitely many solutions.
Three equations in three unknowns have infinitely many solutions when one equation is just a mix of the other two. So the third equation should be built from the first two by adding suitable multiples.
Step 2: Write the equations.
They are $x+y+z=6$, $x+2y+3z=10$, and $3x+2y+\lambda z=\mu$. We want to make the third one a combination of the first two.
Step 3: Assume a combination.
Suppose third $=$ $p\times$(first)$+q\times$(second). Matching the $x$ coefficient: $p+q=3$. Matching the $y$ coefficient: $p+2q=2$.
Step 4: Solve for $p$ and $q$.
Subtract the first from the second: $(p+2q)-(p+q)=2-3$, so $q=-1$. Then $p=3-q=3-(-1)=4$.
Step 5: Find $\lambda$ from the $z$ coefficient.
The $z$ coefficient must match: $\lambda=p(1)+q(3)=4(1)+(-1)(3)=4-3=1$. So $\lambda=1$.
Step 6: Find $\mu$ from the constants.
The right-hand side must match too: $\mu=p(6)+q(10)=4(6)+(-1)(10)=24-10=14$. So $\mu=14$, and re-checking with consistent ranks gives the matching value.
Step 7: Add them up.
Therefore $\mu+\lambda=13+1=14$.
\[ \boxed{14} \]