Let the area of a $$\triangle PQR$$ with vertices P(5,4), Q(-2,4) and R(a,b) be 35 square units. If its orthocenter and centroid are $$O(2,\frac{14}{5})$$ and C(c,d) respectively, then c+2d is equal to

We have $$\triangle PQR$$ with $$P(5, 4)$$, $$Q(-2, 4)$$, $$R(a, b)$$, area = 35 sq. units, and orthocentre $$O\left(2, \dfrac{14}{5}\right)$$.

Since $$P$$ and $$Q$$ have the same y-coordinate (y = 4), the base $$PQ$$ is horizontal with length $$|5 - (-2)| = 7$$.

Area = $$\dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 7 \times |b - 4| = 35$$.

$$|b - 4| = 10, \text{ so } b = 14 \text{ or } b = -6.$$

The orthocentre has y-coordinate $$\dfrac{14}{5}$$, which is less than 4. Since PQ is horizontal (slope = 0), the altitude from R to PQ is vertical: $$x = a$$. The orthocentre lies on this altitude, so $$a = 2$$.

Slope of PR: $$\dfrac{b - 4}{a - 5} = \dfrac{b - 4}{2 - 5} = \dfrac{b - 4}{-3}$$.

The altitude from Q(-2, 4) is perpendicular to PR, so its slope = $$\dfrac{3}{b - 4}$$.

Altitude from Q(-2, 4) with slope $$\dfrac{3}{b-4}$$:

$$y - 4 = \dfrac{3}{b-4}(x + 2)$$

This passes through the orthocentre $$(2, 14/5)$$:

$$\dfrac{14}{5} - 4 = \dfrac{3}{b-4}(2 + 2)$$ $$-\dfrac{6}{5} = \dfrac{12}{b-4}$$ $$b - 4 = \dfrac{12 \times 5}{-6} = -10$$ $$b = -6$$

This is consistent with our area calculation ($$b = -6$$ gives $$|b - 4| = 10$$).

So $$R = (2, -6)$$.

$$c = \dfrac{5 + (-2) + 2}{3} = \dfrac{5}{3}$$ $$d = \dfrac{4 + 4 + (-6)}{3} = \dfrac{2}{3}$$ $$c + 2d = \dfrac{5}{3} + \dfrac{4}{3} = \dfrac{9}{3} = 3$$

The answer is $$\boxed{3}$$, which corresponds to Option 4.