If the area of the triangle formed by the $$x$$-axis, the normal and the tangent to the circle $$(x - 2)^2 + (y - 3)^2 = 25$$ at the point (5, 7) is $$A$$, then $$24A$$ is equal to ______.

The circle is $$(x - 2)^2 + (y - 3)^2 = 25$$ with center $$C = (2, 3)$$. The given point is $$(5, 7)$$, which lies on the circle since $$(5-2)^2 + (7-3)^2 = 9 + 16 = 25$$.

The slope of the radius from the center $$(2, 3)$$ to $$(5, 7)$$ is $$\frac{7 - 3}{5 - 2} = \frac{4}{3}$$. The normal to the circle at $$(5, 7)$$ passes through the center and has slope $$\frac{4}{3}$$, so its equation is $$y - 7 = \frac{4}{3}(x - 5)$$. Setting $$y = 0$$: $$-7 = \frac{4}{3}(x - 5)$$, giving $$x = 5 - \frac{21}{4} = -\frac{1}{4}$$. The normal meets the $$x$$-axis at $$\left(-\frac{1}{4}, 0\right)$$.

The tangent at $$(5, 7)$$ is perpendicular to the radius, so it has slope $$-\frac{3}{4}$$. Its equation is $$y - 7 = -\frac{3}{4}(x - 5)$$. Setting $$y = 0$$: $$-7 = -\frac{3}{4}(x - 5)$$, giving $$x = 5 + \frac{28}{3} = \frac{43}{3}$$. The tangent meets the $$x$$-axis at $$\left(\frac{43}{3}, 0\right)$$.

The triangle is formed by the point $$(5, 7)$$, the $$x$$-intercept of the normal $$\left(-\frac{1}{4}, 0\right)$$, and the $$x$$-intercept of the tangent $$\left(\frac{43}{3}, 0\right)$$. The base along the $$x$$-axis has length $$\frac{43}{3} - \left(-\frac{1}{4}\right) = \frac{43}{3} + \frac{1}{4} = \frac{172 + 3}{12} = \frac{175}{12}$$, and the height is $$7$$.

The area is $$A = \frac{1}{2} \times \frac{175}{12} \times 7 = \frac{1225}{24}$$. Therefore, $$24A = 1225$$.

The correct answer is $$1225$$.