Let the tangent to the parabola $$S : y^2 = 2x$$ at the point $$P(2, 2)$$ meet the $$x$$-axis at $$Q$$ and normal at it meet the parabola $$S$$ at the point $$R$$. Then the area (in sq. units) of the triangle $$PQR$$ is equal to:

The parabola is $$y^2 = 2x$$. Differentiating: $$2y\frac{dy}{dx} = 2$$, so the slope at $$P(2,2)$$ is $$\frac{dy}{dx} = \frac{1}{y} = \frac{1}{2}$$.

Equation of the tangent at $$P(2,2)$$: $$y - 2 = \frac{1}{2}(x-2)$$, which simplifies to $$x - 2y + 2 = 0$$. Setting $$y = 0$$ gives $$x = -2$$, so $$Q = (-2, 0)$$.

The slope of the normal at $$P$$ is $$-2$$ (negative reciprocal of the tangent slope). Equation of the normal: $$y - 2 = -2(x - 2)$$, i.e., $$y = -2x + 6$$.

To find where the normal meets the parabola again, substitute $$y = -2x+6$$ into $$y^2 = 2x$$: $$(-2x+6)^2 = 2x \implies 4x^2 - 24x + 36 = 2x \implies 4x^2 - 26x + 36 = 0 \implies 2x^2 - 13x + 18 = 0.$$

The roots are $$x = \frac{13 \pm \sqrt{169 - 144}}{4} = \frac{13 \pm 5}{4}$$, giving $$x = 2$$ (point $$P$$) and $$x = \frac{9}{2}$$. At $$x = \frac{9}{2}$$: $$y = -2\cdot\frac{9}{2} + 6 = -3$$. So $$R = \left(\frac{9}{2}, -3\right)$$.

Area of triangle $$PQR$$ with $$P(2,2)$$, $$Q(-2,0)$$, $$R\!\left(\tfrac{9}{2},-3\right)$$: Area $$= \frac{1}{2}\left|x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)\right|$$ $$= \frac{1}{2}\left|2(0-(-3)) + (-2)((-3)-2) + \frac{9}{2}(2-0)\right|$$ $$= \frac{1}{2}\left|6 + 10 + 9\right| = \frac{25}{2}.$$

The area of triangle $$PQR$$ is $$\dfrac{25}{2}$$ square units.