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Need help on a vector question!

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There are 7 cases where 3 planes can intersect.
Trivial cases: 1) three planes identical, plane of solutions 2) two planes identical third plane different normal: line of sol'n 3) two planes identical third plane is parallel, no solution 4) all three planes parallel non intersecting, also no solution

Non-trivial cases
Where the triple scalar product of the 3 normals = 0 (indicating the 3 distinct normals are coplanar and each is linearly dependent of the other two
5) If infinite solution, the line intersection of 2 planes is also on the third plane (tsp and this are two equations).
6) If no solution, the line intersection of 2 planes does not lie on the third.

Where tsp =/= 0, the three normals form a basis for a 3-d space
7) Use tsp to solve for alpha. Plug the line intersection of known equations to solve for beta.

Alternatively, if your school taught you about reduced echelon form, 
for 5) you get a row of all 0s in the augmented matrix 6) you get inconsistency (0 = non-zero number) 7) you get variable = value (eg z = 4)  

I got 
a) i) alpha = 2, beta =/= 0
ii) alpha =/= 2, beta is real (any real because plane 3 cannot be same or parallel to plane 1 or 2)
iii) alpha = 2, beta = 0
b) EDIT: gif.latex?\frac{2-x}{2}=\frac{4-y}{2}=z Initally I put vector form but I didn't read question correctly
I solved it using RREF (reduced row echelon form) but if you are unable to solve it using triple scalar product and plug in the equation of a line let me know.
EDIT: I was able to get the answer without using RREF or matrix operations.
Note the point (-2, 4, 0) is the point you would get using RREF. If not you can use any other point on the line but anyhow (-2, -2, 1) is the direction vector.

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