Question 1198864
<br>
Solve the 3rd equation for z and substitute in the other two equations.<br>
x = 3z<br>
3kz+y+z=0
3z+2y+kz=0<br>
y+(3k+1)z=0
2y+(3+k)z=0<br>
There are infinitely many solutions if the two equations are equivalent.<br>
2y+(6k+2)z=0
2y+(3+k)z=0<br>
6k+2=3+k
5k=1
k=1/5<br>
ANSWER: k=1/5<br>