Let a ray of light passing through the point $$(3, 10)$$ reflects on the line $$2x + y = 6$$ and the reflected ray passes through the point $$(7, 2)$$. If the equation of the incident ray is $$ax + by + 1 = 0$$, then $$a^2 + b^2 + 3ab$$ is equal to _____

A ray of light passes through the point $$(3, 10)$$, reflects on the line $$2x + y = 6$$, and the reflected ray passes through $$(7, 2)$$.

Find the image of $$(3, 10)$$ in the line $$2x + y = 6$$.

The line is $$2x + y - 6 = 0$$. The image of point $$(x_1, y_1)$$ in line $$ax + by + c = 0$$ is:

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$

For $$(3, 10)$$ in $$2x + y - 6 = 0$$:

$$\frac{x - 3}{2} = \frac{y - 10}{1} = \frac{-2(6 + 10 - 6)}{4 + 1} = \frac{-2(10)}{5} = -4$$

$$x = 3 + 2(-4) = -5, \quad y = 10 + 1(-4) = 6$$

The image is $$(-5, 6)$$.

The reflected ray passes through $$(7, 2)$$ and the image $$(-5, 6)$$.

But actually, for reflection, the reflected ray from $$(7, 2)$$ appears to come from the image of the source. So the incident ray from $$(3, 10)$$ hits the mirror and reflects to $$(7, 2)$$. The reflected ray, when extended, appears to come from the image of $$(3, 10)$$.

So the reflected ray passes through $$(-5, 6)$$ and $$(7, 2)$$.

The point of incidence $$P$$ is where this line meets $$2x + y = 6$$.

Line through $$(-5, 6)$$ and $$(7, 2)$$:

$$\frac{y - 6}{x + 5} = \frac{2 - 6}{7 + 5} = \frac{-4}{12} = -\frac{1}{3}$$

$$y - 6 = -\frac{1}{3}(x + 5)$$

$$3y - 18 = -x - 5$$

$$x + 3y = 13$$

Find intersection with $$2x + y = 6$$:

From $$x + 3y = 13$$: $$x = 13 - 3y$$

$$2(13 - 3y) + y = 6 \Rightarrow 26 - 6y + y = 6 \Rightarrow -5y = -20 \Rightarrow y = 4$$

$$x = 13 - 12 = 1$$

Point of incidence is $$P(1, 4)$$.

Find the equation of the incident ray through $$(3, 10)$$ and $$(1, 4)$$.

Slope $$= \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3$$

$$y - 4 = 3(x - 1) \Rightarrow y = 3x + 1 \Rightarrow 3x - y + 1 = 0$$

In the form $$ax + by + 1 = 0$$: $$3x - y + 1 = 0$$ gives $$a = 3, b = -1$$.

Compute $$a^2 + b^2 + 3ab$$.

$$a^2 + b^2 + 3ab = 9 + 1 + 3(3)(-1) = 10 - 9 = 1$$

The answer is 1.