Points $$P(-3, 2)$$, $$Q(9, 10)$$ and $$R(a, 4)$$ lie on a circle $$C$$ with $$PR$$ as its diameter. The tangents to $$C$$ at the points $$Q$$ and $$R$$ intersect at the point $$S$$. If $$S$$ lies on the line $$2x - ky = 1$$, then $$k$$ is equal to _____.

Points $$P(-3, 2)$$, $$Q(9, 10)$$, and $$R(\alpha, 4)$$ lie on a circle $$C$$ with $$PR$$ as its diameter.

Find the center and radius of circle $$C$$.

Since $$PR$$ is the diameter, the center is the midpoint of $$PR$$:

$$ \text{Center} = \left(\frac{-3 + \alpha}{2}, \frac{2 + 4}{2}\right) = \left(\frac{\alpha - 3}{2}, 3\right) $$

Use the condition that $$Q(9, 10)$$ lies on the circle.

The radius equals both the distance from center to $$P$$ and to $$Q$$:

$$ \left(\frac{\alpha - 3}{2} + 3\right)^2 + (3-2)^2 = \left(\frac{\alpha - 3}{2} - 9\right)^2 + (3-10)^2 $$ $$ \left(\frac{\alpha + 3}{2}\right)^2 + 1 = \left(\frac{\alpha - 21}{2}\right)^2 + 49 $$

Expanding:

$$ \frac{(\alpha + 3)^2}{4} + 1 = \frac{(\alpha - 21)^2}{4} + 49 $$ $$ (\alpha + 3)^2 + 4 = (\alpha - 21)^2 + 196 $$ $$ \alpha^2 + 6\alpha + 9 + 4 = \alpha^2 - 42\alpha + 441 + 196 $$ $$ 48\alpha = 624 \implies \alpha = 13 $$

So $$R = (13, 4)$$, Center = $$(5, 3)$$, Radius$$^2 = (5+3)^2 + (3-2)^2 = 64 + 1 = 65$$.

Find the tangent at $$Q(9, 10)$$.

$$ (9-5)(x-5) + (10-3)(y-3) = 65 $$ $$ 4x - 20 + 7y - 21 = 65 \implies 4x + 7y = 106 $$

Find the tangent at $$R(13, 4)$$.

$$ (13-5)(x-5) + (4-3)(y-3) = 65 $$ $$ 8x - 40 + y - 3 = 65 \implies 8x + y = 108 $$

Find the intersection point $$S$$.

From $$8x + y = 108$$: $$y = 108 - 8x$$

Substituting into $$4x + 7y = 106$$:

$$ 4x + 7(108 - 8x) = 106 $$ $$ 4x + 756 - 56x = 106 $$ $$ -52x = -650 \implies x = \frac{325}{26} = \frac{25}{2} $$ $$ y = 108 - 8 \cdot \frac{25}{2} = 108 - 100 = 8 $$

So $$S = \left(\frac{25}{2}, 8\right)$$.

Find $$k$$.

$$S$$ lies on $$2x - ky = 1$$:

$$ 2 \cdot \frac{25}{2} - k \cdot 8 = 1 $$ $$ 25 - 8k = 1 \implies k = 3 $$

The value of $$k$$ is $$\boxed{3}$$.