The equations are
If , so is . Then from the last equation or . These are 2 points.
If , then you get or and or . All four combinations are valid, so you get another 4 points.
The last option is if none of them are 0. Then \lambda is non-zero also because if it was, then we'd get one of . If you solve this you should get 4 solutions where , , (where some of the signs change, i.e. , , )
This is a grand total of 10 solutions. The way you did it is incorrect.