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(C) To find the reflection of point $B(7,2)$ about the line $2x+y-6=0$,let the reflected point be $B'(x', y')$.Using the reflection formula $\frac{x'-

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(C) To find the reflection of point $B(7,2)$ about the line $2x+y-6=0$,let the reflected point be $B'(x', y')$.Using the reflection formula $\frac{x'-7}{2} = \frac{y'-2}{1} = -2 \left( \frac{2(7)+1(2)-6}{2^2+1^2} \right)$,$\frac{x'-7}{2} = \frac{y'-2}{1} = -2 \left( \frac{14+2-6}{5} \right) = -2 \left( \frac{10}{5} \right) = -4$.Thus,$x'-7 = -8 \implies x' = -1$ and $y'-2 = -4 \implies y' = -2$.So,$B' = (-1, -2)$.The incident ray passes through $A(3, 10)$ and $B'(-1, -2)$.The slope of the incident ray is $m = \frac{10 - (-2)}{3 - (-1)} = \frac{12}{4} = 3$.The equation of the incident ray is $y - 10 = 3(x - 3) \implies y - 10 = 3x - 9 \implies 3x - y + 1 = 0$.Comparing this with $ax + by + 1 = 0$,we get $a = 3$ and $b = -1$.Therefore,$a^2 + b^2 + 3ab = (3)^2 + (-1)^2 + 3(3)(-1) = 9 + 1 - 9 = 1$.

Let a ray of light passing through the point $(3,10)$ reflect on the line $2x+y=6$ and the reflected ray pass 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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