1. Write the system in matrix form:

Coefficient matrix:

For the system to have no solution, we need:

• $\det(A)=0$ (so the equations are dependent/inconsistent possible), and
• the augmented system must be inconsistent.

2. Compute the determinant:

Expanding along the first row,

So,

3. Substitute $\lambda=2$ into the third equation:

Now compare with the first two equations.

From the first equation,

Multiply by $2$:

Subtract the third equation from this relation is not immediately helpful, so instead solve dependency directly.

Take equation (2) minus $3\times$ equation (1):

Using equation (1):

Then the left side of equation (3) with $\lambda=2$ becomes

Thus equation (3) becomes

• If $\mu=10$, the third equation is consistent with the first two, so infinitely many solutions exist.
• If $\mu\ne 10$, the third equation contradicts the first two, so there is no solution.

4. Therefore, for no solution:

5. Check options:

• A: $\lambda=3,\mu=5$ → $\det(A)\ne 0$, unique solution, not correct.
• B: $\lambda=3,\mu\ne 10$ → still $\det(A)\ne 0$, not correct.
• C: $\lambda\ne 2,\mu=10$ → generally unique solution, not correct.
• D: $\lambda=2,\mu\ne 10$ → gives no solution, correct.

Hence the correct option is D.