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

Solution

1. Consider the system of equations: $x+y+z=6$, $x+2y+3z=10$, and $x+2y+kz=m$.
2. For the system to have no solution, the second and third equations must represent parallel planes, which requires the coefficients of $x,y,$ and $z$ to be proportional, implying $k=3$.
3. With $k=3$, the second and third equations become $x+2y+3z=10$ and $x+2y+3z=m$.
4. If $m\ne 10$, these two planes are parallel and distinct, meaning there is no point of intersection for the system.
5. Hence the answer is (B).