The normal to the plane has the direction (3,−1,4).
Therefore, the basis must consist of two linearly independent vectors that are orthogonal to it.
3a1−b1+4c1=0
3a2−b2+4c2=0 We consider the set {(0,4,1),(1,3,0)}.
We can see that for any scalar k,
(0,4,1)=k(1,3,0)Therefore, they are linearly independent.
Also,
(0,4,1)⋅(3,−1,4)=0
(1,3,0)⋅(3,−1,4)=0
Therefore, the basis is {(0,4,1),(1,3,0)}.
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