Given equation of the plane: x - 2y + 3z = 0
Let y = t and z = s, then x = 2t - 3s
So, the basis for the plane can be written as:
v1 = (2, 1, 0)
v2 = (-3, 0, 1)
Therefore, the basis for the plane x-2y+3z=0 in R^3 is { (2, 1, 0), (-3, 0, 1) }.
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