a)
N = [-2, 1, 1] ⨯ [3, 0, -2] = [-2, -1, -3] = - [2, 1, 3]
E1: (X - [0, 1, 1])·[2, 1, 3]/√14 = 0
E2: (X - [1, 1, 1])·[1, 1, 2]/√6 = 0
E2: X = [6, 0, 0] + r·[-6, 6, 0] + s·[-6, 0, 3]
b)
2·x + y + 3·z = 4
x + y + 2·z = 6
s: X = [0, 10, -2] + r·[1, 1, -1]
c)
2·(3 + t) + (1 + t) + 3·(-1 - t) = 4 → immer wahr, damit liegt die Gerade in der Ebene.