The number of values of a ∈ N such that the variance of 3, 7, 12,...

Calculating Mean $= \dfrac{3 + 7 + 12 + a + 43 - a}{5} = 13$

Variance $= \dfrac{9 + 49 + 144 + {a}^{2} + \left(43 - a\right){}^{2}}{5} - {13}^{2} \in N$

$⇒ \dfrac{2{a}^{2} - a + 1}{5} = n$ (where $n \in N$)

$⇒ 2{a}^{2} - a + 1 - 5n = 0$ must have solution as natural numbers

So, $D = 40n - 7$ always has $3$ at unit place for any value of $n$

$⇒ D$ can't be perfect square

So, $a$ can't be integer.