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(D) Let the point $P = (\lambda^2 + 1, \lambda)$ and its image $P' = (\lambda, \lambda - 1)$ with respect to the line $3x + y - 6k = 0$.$1$. The midpo

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$2$

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(D) Let the point $P = (\lambda^2 + 1, \lambda)$ and its image $P' = (\lambda, \lambda - 1)$ with respect to the line $3x + y - 6k = 0$.$1$. The midpoint of $PP'$ must lie on the line $3x + y - 6k = 0$.Midpoint $M = (\frac{\lambda^2 + 1 + \lambda}{2}, \frac{\lambda + \lambda - 1}{2}) = (\frac{\lambda^2 + \lambda + 1}{2}, \frac{2\lambda - 1}{2})$.Substituting $M$ into the line equation: $3(\frac{\lambda^2 + \lambda + 1}{2}) + (\frac{2\lambda - 1}{2}) = 6k$.$3\lambda^2 + 3\lambda + 3 + 2\lambda - 1 = 12k \implies 3\lambda^2 + 5\lambda + 2 = 12k$.$2$. The line $PP'$ must be perpendicular to the line $3x + y - 6k = 0$.The slope of $PP'$ is $m_{PP'} = \frac{(\lambda - 1) - \lambda}{(\lambda) - (\lambda^2 + 1)} = \frac{-1}{-\lambda^2 + \lambda - 1} = \frac{1}{\lambda^2 - \lambda + 1}$.The slope of the line $3x + y - 6k = 0$ is $m_L = -3$.Since $m_{PP'} 	imes m_L = -1$,we have $(\frac{1}{\lambda^2 - \lambda + 1}) 	imes (-3) = -1 \implies \lambda^2 - \lambda + 1 = 3 \implies \lambda^2 - \lambda - 2 = 0$.Solving for $\lambda$: $(\lambda - 2)(\lambda + 1) = 0$,so $\lambda = 2$ or $\lambda = -1$.$3$. Calculate $k$ for each $\lambda$:If $\lambda = 2$: $12k = 3(2)^2 + 5(2) + 2 = 12 + 10 + 2 = 24 \implies k = 2$.If $\lambda = -1$: $12k = 3(-1)^2 + 5(-1) + 2 = 3 - 5 + 2 = 0 \implies k = 0$.Sum of values of $k = 2 + 0 = 2$.

Sum of values of $k$ for which the image of the point $(\lambda^2 + 1, \lambda)$ with respect to the line $y = -3x + 6k$ is the point $(\lambda, \lambda - 1)$ is:

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