Let the point $$P(\alpha, \beta)$$ be at a unit distance from each of the two lines $$L_1: 3x - 4y + 12 = 0$$, and $$L_2: 8x + 6y + 11 = 0$$. If $$P$$ lies below $$L_1$$ and above $$L_2$$, then $$100(\alpha + \beta)$$ is equal to

We wish to determine the value of $$100(\alpha + \beta)$$ for the point $$P(\alpha,\beta)$$ which lies at a unit distance from both lines, below $$L_1$$ and above $$L_2$$.

Consider the two lines $$L_1: 3x - 4y + 12 = 0$$ and $$L_2: 8x + 6y + 11 = 0$$.

Because the coefficient of $$y$$ in $$L_1$$ is negative, a point below this line makes the expression $$3x - 4y + 12$$ positive. Testing the point $$(0,-100)$$ indeed gives $$3(0) - 4(-100) + 12 = 412 > 0$$, so “below $$L_1$$” corresponds to $$3\alpha - 4\beta + 12 > 0$$. Since the distance from $$P$$ to $$L_1$$ is 1, we have

$$\frac{3\alpha - 4\beta + 12}{\sqrt{3^2 + (-4)^2}} = 1 \quad\Longrightarrow\quad \frac{3\alpha - 4\beta + 12}{5} = 1 \quad\Longrightarrow\quad 3\alpha - 4\beta = -7.$$

For $$L_2$$ a point above it makes the expression $$8x + 6y + 11$$ positive. Checking $$(0,100)$$ yields $$8(0) + 6(100) + 11 = 611 > 0$$, so “above $$L_2$$” means $$8\alpha + 6\beta + 11 > 0$$. Requiring the distance to be 1 gives

$$\frac{8\alpha + 6\beta + 11}{\sqrt{8^2 + 6^2}} = 1 \quad\Longrightarrow\quad \frac{8\alpha + 6\beta + 11}{10} = 1 \quad\Longrightarrow\quad 8\alpha + 6\beta = -1.$$

Thus we obtain the linear system $$3\alpha - 4\beta = -7\quad\text{and}\quad 8\alpha + 6\beta = -1.$$ Solving the first for $$\alpha$$ gives $$\alpha = \frac{-7 + 4\beta}{3}$$. Substituting into the second equation leads to

Substituting back into $$\alpha = \frac{-7 + 4\beta}{3}$$ yields

$$\alpha = \frac{-7 + 4 \times \frac{53}{50}}{3} = \frac{-7 + \frac{212}{50}}{3} = \frac{\frac{-350 + 212}{50}}{3} = \frac{-138}{150} = -\frac{23}{25}.$$

Therefore

$$\alpha + \beta = -\frac{23}{25} + \frac{53}{50} = -\frac{46}{50} + \frac{53}{50} = \frac{7}{50},$$

and hence

$$100(\alpha + \beta) = 100 \times \frac{7}{50} = 14.$$

The correct answer is Option D: $$14$$.