Concept:
Two planes are parallel if their normal vectors are parallel.
So if
\[
\vec{n}_1=(2,-\lambda,1)
\quad \text{and} \quad
\vec{n}_2=(4,-1,\mu),
\]
then for parallel planes:
\[
(2,-\lambda,1)=k(4,-1,\mu)
\]
for some scalar \(k\).
ip
Step 1: Equate the corresponding components.
From the first component:
\[
2 = 4k
\]
\[
k = \frac{1}{2}
\]
ip
Step 2: Find \(\lambda\).
From the second component:
\[
-\lambda = -1 \cdot \frac{1}{2}
\]
\[
-\lambda = -\frac{1}{2}
\]
\[
\lambda = \frac{1}{2}
\]
ip
Step 3: Find \(\mu\).
From the third component:
\[
1 = \mu \cdot \frac{1}{2}
\]
\[
\mu = 2
\]
ip
Step 4: Calculate \(\lambda+\mu\).
\[
\lambda+\mu = \frac{1}{2}+2 = \frac{5}{2}
\]
ip
Hence, the correct answer is:
\[
\boxed{(C)\ \frac{5}{2}}
\]