Step 1: Simplify the first system.
The three equations $x+y-2z=0$, $4x+4y-8z=0$, $3x+3y-6z=0$ are all multiples of the single plane $x+y-2z=0$. So the first system is really just one equation.
Step 2: Match it with the second system.
For both systems to share the same solution set, the second system's solutions must satisfy $x+y-2z=0$. Use the second system's first equation $x+y+z=3$ alongside it.
Step 3: Solve for $z$.
Subtract: $(x+y+z) - (x+y-2z) = 3 - 0$, giving $3z = 3$, so $z = 1$.
Step 4: Solve for $x+y$.
Put $z=1$ into $x+y-2z=0$: $x+y-2 = 0$, so $x+y = 2$.
Step 5: Find $\lambda$.
The equation $2x+2y-z=\lambda$ becomes $2(x+y) - z = 2(2) - 1 = 3$, so $\lambda = 3$.
Step 6: Find $\mu$ and add.
The equation $x+y-\mu z=1$ becomes $2 - \mu = 1$, so $\mu = 1$. Therefore $\lambda + \mu = 3 + 1 = 4$, which is option (D).
\[ \boxed{\lambda + \mu = 4} \]