Let S = {(-1, 0, 1), (2, 1, 4)}. The value of k for which the vector (3k + 2, 3,10) belongs to the linear span of S is: 

The correct answer is option 3

Concept:

if vector (3k + 2, 3,10) belongs to the linear span of S then the determinant of vectors is zero.

Calculation:

\(\left| {\begin{array}{*{20}{c}} -1&0&1\\ 2&1&{ 4}\\ 3k+2&3&10 \end{array}} \right| = 0\)

⇒ -1 × (1 × 10 - 4 × 3) - 0 × (2 × 10 - 4 × (3k + 2)) + 1 × (2 × 3 - 1 × (3k + 2)) = 0

⇒ 2 - 0 - 3k + 4 = 0

⇒ 3k = 6

∴ k = 2.

Therefore the value of k for which the vector (3k + 2, 3,10) belongs to the linear span of S is 2.