Let the third vertex of the triangle is C(x, y)

Given, two vertices of a triangle are A(3,-2) and B(-2,3)

Now given orthocentre of the circle = H(-6, 1)

So, AH ⊥ BC and BH ⊥ AC

Since the product of the slope of perpendicular lines equal to -1

Now, AH ⊥ BC

=> {(-2 - 1)/(3 + 6)} * {(y + 2)/(x - 3)} = -1

=> (-3/9) * {(y + 2)/(x - 3)} = -1

=> (-1/3)*{(y - 3)/(x + 2)} = -1

=> (y - 3)/{3*(x + 2)} = 1

=> (y - 3) = 3*(x + 2)

=> y - 3 = 3x + 6

=> 3x + 6 - y = -3

=> 3x - y = -3 - 6

=> 3x - 2y = -9 ............1

Again, BH ⊥ AC

=> {(3 - 1)/(-2 + 6)} * {(y - 3)/(x + 2)} = -1

=> (2/4) * {(y - 3)/(x + 2)} = -1

=> (1/2)*{(y - 3)/(x + 2)} = -1

=> (y - 3)/{2*(x + 2)} = 1

=> (y - 3) = 2*(x + 2)

=> y - 3 = 2x + 4

=> 2x + 4 - y = -3

=> 2x - y = -3 - 4

=> 2x - y = -7 ............2

 Multiply equation 2 by 2, we get

4x - 2y = -14 .........3

Subtract equation 1 and , we get

    -x = 5

=> x = -5

From equation 2, we get

    2*(-5) - y = -7

=> -10 - y = -7

=> y = -10 + 7

=> y = -3

So, the third vertex of the triangle is (-5, -3)