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(D) Given the curve $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$. First, find the derivative $\frac{dy}{dx}$:$\frac{dx}{dt} = 2t - 7$ and $\frac{dy}{dt}

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$5$

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(D) Given the curve $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$. First, find the derivative $\frac{dy}{dx}$:$\frac{dx}{dt} = 2t - 7$ and $\frac{dy}{dt} = 2t - 4$.Thus, $\frac{dy}{dx} = \frac{2t - 4}{2t - 7}$. At the point $(1, 2)$, we have $t^2 - 7t + 7 = 1 \implies t^2 - 7t + 6 = 0 \implies (t-1)(t-6) = 0$, so $t = 1$ or $t = 6$. Also, $t^2 - 4t - 10 = 2 \implies t^2 - 4t - 12 = 0 \implies (t-6)(t+2) = 0$, so $t = 6$ or $t = -2$. The common value is $t = 6$. The slope of the tangent at $t = 6$ is $\frac{2(6) - 4}{2(6) - 7} = \frac{8}{5}$. The slope of the normal $m$ is $-\frac{1}{8/5} = -\frac{5}{8}$. The equation of the normal at $(1, 2)$ is $y - 2 = -\frac{5}{8}(x - 1)$. $8y - 16 = -5x + 5 \implies 5x + 8y - 21 = 0$. Here $a = 5, b = 8, c = -21$. The $G$.$C$.$D$. of $(5, 8, -21)$ is $1$. Then $a + b + c = 5 + 8 - 21 = -8$. Finally, $m(a + b + c) = (-\frac{5}{8})(-8) = 5$.

Let $m$ be the slope of the normal $L$ drawn at $(1, 2)$ to the curve $x = t^2 - 7t + 7, y = t^2 - 4t - 10$ and $ax + by + c = 0$ be the equation of the normal $L$. If the $G$.$C$.$D$. of $(a, b, c)$ is $1$, then $m(a + b + c) =$

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