Question
If the distance of the point $(a,2,5)$ from the image of the point $(1,2,7)$ in the line $\frac{x}{1} = \frac{y - 1}{1} = \frac{z - 2}{2}$ is 4, then the sum of all possible values of a is equal to :
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Solution
Sol. [Key Concept: Reflect $(1,2,7)$ in the line to get its image, set that image's distance to $(a,2,5)$ equal to $4$, and sum the roots of the resulting quadratic in $a$.]
$$PQ\bot L \Rightarrow (\alpha - 1) + (\beta - 2) + 2(\gamma - 7) = 0$$
$$\begin{array}{r} \Rightarrow \alpha + \beta + 2\gamma = 17\#(1) \end{array}$$
M is mid point of PQ which will satisfy L
$${\frac{\alpha + 1}{2} = \frac{\frac{\beta + 2}{2} - 1}{1} = \frac{\frac{\gamma + 7}{2} - 2}{2} }{\Rightarrow \frac{\alpha + 1}{2} = \frac{\beta}{2} = \frac{\gamma + 3}{4}}$$
$$\begin{array}{r} \Rightarrow \alpha + 1 = \beta\#(2) \end{array}$$
and $2\beta = \gamma + 3$
$$\begin{array}{r} \Rightarrow \alpha = 3,\beta = 4,\gamma = 5\#(3) \end{array}$$
Distance from (a, 2, 5) is $= \sqrt{(a - 3)^{2} + 4 + 0} = 4$
$(a - 3)^{2} + 4 = 16 \Rightarrow a^{2} - 6a - 3 = 0 \Rightarrow$ sum of values of $a = 6$
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