A ray of light coming from the point $$P(1, 2)$$ gets reflected from the point $$Q$$ on the $$x$$-axis and then passes through the point $$R(4, 3)$$. If the point $$S(h, k)$$ is such that PQRS is a parallelogram, then $$hk^2$$ is equal to :

A ray from $$P(1,2)$$ reflects off the x-axis at point Q and passes through $$R(4,3)$$.

When reflecting off the x-axis, the image of P in the x-axis is $$P'(1, -2)$$. The reflected ray passes through $$P'$$ and $$R$$.

Line through $$P'(1,-2)$$ and $$R(4,3)$$: slope = $$\frac{3-(-2)}{4-1} = \frac{5}{3}$$.

Equation: $$y + 2 = \frac{5}{3}(x-1) \implies 3y + 6 = 5x - 5 \implies 5x - 3y = 11$$.

Point Q (on x-axis, y = 0): $$5x = 11 \implies x = \frac{11}{5}$$. So $$Q = \left(\frac{11}{5}, 0\right)$$.

PQRS is a parallelogram, so $$\vec{QR} = \vec{PS}$$.

$$\vec{QR} = R - Q = \left(4 - \frac{11}{5}, 3 - 0\right) = \left(\frac{9}{5}, 3\right)$$.

$$S = P + \vec{QR} = \left(1 + \frac{9}{5}, 2 + 3\right) = \left(\frac{14}{5}, 5\right)$$.

So $$h = \frac{14}{5}$$, $$k = 5$$.

$$hk^2 = \frac{14}{5} \times 25 = 14 \times 5 = 70$$.

The correct answer is Option 1: 70.