If G(−2, 1) is the centroid of a ∆ABC and two of its vertices are A(1, −6) and B(−5, 2),

Two vertices of ∆ABC are A(1, −6) and B(−5, 2). Let the third vertex be C(a, b).
Then the coordinates of its centroid are

$C\left(\frac{1-5+a}{3}, \frac{-6+2+b}{3}\right)$

$C\left(\frac{-4+a}{3}, \frac{-4+b}{3}\right)$

But it is given that G(−2, 1) is the centroid. Therefore,

$-2=\frac{-4+a}{3}, 1=\frac{-4+b}{3}$

$\Rightarrow-6=-4+a, 3=-4+b$

$\Rightarrow-6+4=a, 3+4=b$

$\Rightarrow a=-2, b=7$

Therefore, the third vertex of ∆ABC is C(−2, 7).