If the system of equations $x+y+z=6$, $x+2y+3z=10$, and $x+2y+\lambda z=\mu$ has no solution, then:

Solution

1. Consider the system of equations: $x+y+z=6$, $x+2y+3z=10$, and $x+2y+\lambda z=\mu$.
2. Subtract the second equation from the third to obtain $(\lambda -3)z=\mu -10$.
3. For the system to have no solution, the coefficient of $z$ must be zero while the constant term is non-zero.
4. Setting $\lambda -3=0$ gives $\lambda =3$, and setting $\mu -10\ne 0$ gives $\mu \ne 10$.
5. Hence the answer is (B).