If $$(\alpha, \beta)$$ is the orthocenter of the triangle $$ABC$$ with vertices $$A(3, -7)$$, $$B(-1, 2)$$ and $$C(4, 5)$$, then $$9\alpha - 6\beta + 60$$ is equal to

Finding the orthocenter of triangle with vertices A(3,-7), B(-1,2), C(4,5).

Slope of BC: $$m_{BC} = \frac{5-2}{4-(-1)} = \frac{3}{5}$$

Altitude from A ⊥ BC: slope = $$-\frac{5}{3}$$

Equation: $$y + 7 = -\frac{5}{3}(x - 3)$$, i.e., $$3y + 21 = -5x + 15$$, i.e., $$5x + 3y = -6$$ ... (1)

Slope of AC: $$m_{AC} = \frac{5-(-7)}{4-3} = 12$$

Altitude from B ⊥ AC: slope = $$-\frac{1}{12}$$

Equation: $$y - 2 = -\frac{1}{12}(x + 1)$$, i.e., $$12y - 24 = -x - 1$$, i.e., $$x + 12y = 23$$ ... (2)

From (1): $$5x + 3y = -6$$

From (2): $$x = 23 - 12y$$

Substituting: $$5(23 - 12y) + 3y = -6$$

$$115 - 60y + 3y = -6$$

$$-57y = -121$$

$$y = \frac{121}{57} = \frac{121}{57}$$

$$x = 23 - 12 \times \frac{121}{57} = 23 - \frac{1452}{57} = \frac{1311 - 1452}{57} = \frac{-141}{57} = \frac{-47}{19}$$

So $$\alpha = -\frac{47}{19}$$, $$\beta = \frac{121}{57} = \frac{121}{57}$$.

$$9\alpha - 6\beta + 60 = 9 \times \frac{-47}{19} - 6 \times \frac{121}{57} + 60$$

$$= \frac{-423}{19} - \frac{726}{57} + 60 = \frac{-423}{19} - \frac{242}{19} + 60 = \frac{-665}{19} + 60 = -35 + 60 = 25$$

This matches option 1: 25.