Given that point (2,k) lies outsides the circles x² + y² + x - 2y - 14 = 0 and x² + y² = 13.

We know that for a point (a, b) to lie outside the circle S, the condition to be satisfies is S₁₁>0.

Applying S₁₁>0 for 1^st circle,

⇒ 2² + k² + 2 - 2(k) - 14>0

⇒ 4 + k² - 2k - 12>0

⇒ k² - 2k - 8>0

⇒ k² - 4k + 2k - 8>0

⇒ k(k - 4) + 2(k - 4)>0

⇒ (k + 2)(k - 4)>0

We know that the solution set of (x - a)(x - b)>0 for b>a is (- ∞,a)∪(b,∞).

The solution set for k is (- ∞, - 2)∪(4,∞) ..... (1)

Applying S₁₁>0 for 2^nd circle,

⇒ 2² + k² - 13>0

⇒ k² + 4 - 13>0

⇒ k² - 9>0

⇒ (k - 3)(k + 3)>0

The solution set for k is (- ∞, - 3)∪(3,∞) ...... (2)

The resultant solution set for k is the intersection of (1) and (2).

⇒ k(((- ∞, - 2)∪(4,∞))∩((- ∞, - 3)∪(3,∞)))

⇒ k((- ∞, - 3)∪(4,∞))

∴The correct option is (c).