Question
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 $Q(5,\text{ }b,c)$, then $a^{2} + b^{2} + c^{2}$ is equal to
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Solution
Point $M \equiv \left( 3,\frac{\text{ }b}{2} + 1,\frac{c + a}{2} \right)$ satisfies the line
$${\frac{3 - 6}{3} = \frac{\frac{b}{2} + 1 - 7}{2} = \frac{\frac{c + a}{2} - 7}{- 2} }{- 1 = \frac{b - 12}{4} = \frac{c + a - 14}{- 4} }$$$\Rightarrow b = 8\ \ldots$ (1) & $c + a = 18\ldots$ (2)
Now $PQ\bot L$
$${\Rightarrow (4i + (b - 2)j + (c - a)k) \cdot (3i + 2j - 2k) = 0 }{\Rightarrow 12 + 2(\text{ }b - 2) - 2(c - a) = 0 }{\Rightarrow 6 + (b - 2) - (c - a) = 0 }{\Rightarrow b - c + a + 4 = 0 }{\Rightarrow 8 - c + a + 4 = 0}$$
$$\begin{array}{r} \Rightarrow c + a = 12\#(3) \end{array}$$
From (2) & (3)
$$c = 15\& a = 3 $$So $a^{2} + b^{2} + c^{2} = 9 + 64 + 225 = 298$
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