The tangent and normal to the ellipse $$3x^2 + 5y^2 = 32$$ at the point P(2, 2) meet the x-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is:

We are given the ellipse $$3x^2 + 5y^2 = 32$$ and the point $$P(2,\,2)$$ on it. First we differentiate the equation of the ellipse implicitly to find the slope of the tangent. Differentiating, we obtain $$\frac{d}{dx}\,(3x^2) + \frac{d}{dx}\,(5y^2) = \frac{d}{dx}\,(32).$$

This gives $$6x + 10y\frac{dy}{dx} = 0.$$

So $$\frac{dy}{dx} = -\,\frac{6x}{10y} = -\,\frac{3x}{5y}.$$

At the point $$P(2,\,2)$$ we substitute $$x = 2,\; y = 2$$ to get the slope of the tangent:

$$m_{\text{tan}} = -\,\frac{3(2)}{5(2)} = -\,\frac{6}{10} = -\,\frac{3}{5}.$$

Now we write the tangent line in point-slope form $$y - y_1 = m(x - x_1).$$ Using $$P(2,\,2)$$ and $$m_{\text{tan}} = -\tfrac{3}{5},$$ we have

$$y - 2 = -\,\frac{3}{5}\,(x - 2).$$

To find point $$Q$$ where this tangent meets the x-axis, we set $$y = 0.$$ Substituting,

$$0 - 2 = -\,\frac{3}{5}\,(x_Q - 2).$$

Hence $$-2 = -\,\frac{3}{5}(x_Q - 2).$$

Multiplying both sides by $$-\frac{5}{3}$$, we get $$x_Q - 2 = \frac{10}{3},$$ so

$$x_Q = 2 + \frac{10}{3} = \frac{16}{3}.$$

Thus $$Q\Bigl(\frac{16}{3},\,0\Bigr).$$

Next, the slope of the normal is the negative reciprocal of the slope of the tangent. Since $$m_{\text{tan}} = -\frac{3}{5},$$ the normal slope is

$$m_{\text{norm}} = \frac{5}{3}.$$

Using the same point-slope form for the normal through $$P(2,\,2),$$ we write

$$y - 2 = \frac{5}{3}\,(x - 2).$$

To locate point $$R$$ where this normal meets the x-axis, we again set $$y = 0.$$ Substituting,

$$0 - 2 = \frac{5}{3}\,(x_R - 2).$$

This simplifies to $$-2 = \frac{5}{3}(x_R - 2).$$

Multiplying both sides by $$\frac{3}{5},$$ we obtain $$x_R - 2 = -\frac{6}{5},$$ hence

$$x_R = 2 - \frac{6}{5} = \frac{10}{5} - \frac{6}{5} = \frac{4}{5}.$$

Thus $$R\Bigl(\frac{4}{5},\,0\Bigr).$$

Now we have the three vertices of triangle $$PQR$$: $$P(2,\,2), \; Q\Bigl(\frac{16}{3},\,0\Bigr), \; R\Bigl(\frac{4}{5},\,0\Bigr).$$

Because $$Q$$ and $$R$$ lie on the x-axis, segment $$QR$$ is horizontal. Its length (the base of the triangle) is the absolute difference of their x-coordinates:

$$|\,x_Q - x_R| = \left|\,\frac{16}{3} - \frac{4}{5}\right|.$$

Using a common denominator $$15,$$ we have $$\frac{16}{3} = \frac{80}{15}, \qquad \frac{4}{5} = \frac{12}{15},$$ so

$$|\,x_Q - x_R| = \left|\frac{80}{15} - \frac{12}{15}\right| = \frac{68}{15}.$$

The height of the triangle is simply the y-coordinate of $$P,$$ which equals $$2.$$

The area formula for a triangle is $$\text{Area} = \tfrac12 \times (\text{base}) \times (\text{height}).$$ Substituting the base $$\frac{68}{15}$$ and the height $$2,$$ we obtain

$$\text{Area} = \frac12 \times \frac{68}{15} \times 2 = \frac{68}{15}.$$

Hence, the correct answer is Option A.