If the image of the point P(1, 2, a) in the l | vedclass

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(C) Let the line be $L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2} = k$. Any point on the line is $R(3k+6, 2k+7, -2k+7)$.Since $R$ is the midpoint of

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$298$

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(C) Let the line be $L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2} = k$. Any point on the line is $R(3k+6, 2k+7, -2k+7)$.Since $R$ is the midpoint of $PQ$, where $P(1, 2, a)$ and $Q(5, b, c)$, we have:$3k+6 = \frac{1+5}{2} = 3 \Rightarrow 3k = -3 \Rightarrow k = -1$.Thus, the midpoint $R$ is $(3(-1)+6, 2(-1)+7, -2(-1)+7) = (3, 5, 9)$.Using the midpoint formula:$\frac{1+5}{2} = 3$ (satisfied),$\frac{2+b}{2} = 5 \Rightarrow 2+b = 10 \Rightarrow b = 8$,$\frac{a+c}{2} = 9 \Rightarrow a+c = 18$.Since $PQ$ is perpendicular to the line $L$ with direction vector $\vec{v} = 3\hat{i} + 2\hat{j} - 2\hat{k}$, the vector $\vec{PQ} = (5-1)\hat{i} + (b-2)\hat{j} + (c-a)\hat{k} = 4\hat{i} + 6\hat{j} + (c-a)\hat{k}$ must be parallel to $\vec{v}$.Thus, $\frac{4}{3} = \frac{6}{2} = \frac{c-a}{-2}$.This implies $3 = \frac{c-a}{-2} \Rightarrow c-a = -6$.Solving $a+c=18$ and $c-a=-6$:Adding the equations: $2c = 12 \Rightarrow c = 6$.Subtracting: $2a = 24 \Rightarrow a = 12$.Finally, $a^2+b^2+c^2 = 12^2 + 8^2 + 6^2 = 144 + 64 + 36 = 244$. Wait, re-evaluating the line equation: $\frac{7-z}{2} = \frac{z-7}{-2}$. The direction vector is $(3, 2, -2)$.Vector $\vec{PQ} = (4, b-2, c-a)$. Since $\vec{PQ} \cdot \vec{v} = 0$, $4(3) + (b-2)(2) + (c-a)(-2) = 0 \Rightarrow 12 + 2b - 4 - 2c + 2a = 0 \Rightarrow 2b + 2a - 2c + 8 = 0 \Rightarrow a + b - c = -4$.With $b=8$, $a-c = -12$. Given $a+c=18$, we get $2a = 6 \Rightarrow a=3$ and $c=15$.$a^2+b^2+c^2 = 3^2 + 8^2 + 15^2 = 9 + 64 + 225 = 298$.

If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-z}{2}$ is $Q(5, b, c)$, then $a^{2}+b^{2}+c^{2}$ is equal to

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