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 _.

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Given: For $ B' $: $$ \frac{x - 7}{2} = \frac{y - 2}{1} = -2 \left( \frac{14 + 2 - 6}{5} \right) $$ $$ \frac{x - 7}{2} = \frac{y - 2}{1} = -4 $$ $$ x = -1, \quad y = -2 \implies B'(-1, -2) $$ Incident ray $ AB' $: $$ M_{AB'} = 3 $$ $$ y + 2 = 3(x + 1) $$ $$ 3x - y + 1 = 0 $$ $$ a = 3, \quad b = -1 $$ $$ a^2 + b^2 + 3ab = 9 + 1 - 9 = 1 $$ Answer: 1

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 _.

Correct Answer: 1

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