Yefim S. answered 10/13/20
Math Tutor with Experience
Vectors linearly dependent if det(A) = 0
WE expand this determinant by last column: -12 + 22k - 12k2 + 2k3 = 0
k3 - 6k2 + 11k - 6 = 0.
k3 - k2 - 5k2 + 5k + 6k - 6 = 0
(k - 1)(k2 - 5k + 6) = 0; (k - 1)(k - 2)(k - 3) = 0
k = 1, k = 2, k = 3