JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the mirror image of the point \(\mathrm{P}(3,4,9)\) in the line \(\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}\) is \((\alpha, \beta, \gamma)\), then \(14(\alpha+\beta+\gamma)\) is :
- A \(102\)
- B \(138\)
- C \(108\)
- D \(132\)
Answer & Solution
Correct Answer
(C) \(108\)
Step-by-step Solution
Detailed explanation
\( \overrightarrow{\text { PN. }} \cdot \vec{b}=0 ? \) \( 3(3 \lambda-2)+2(2 \lambda-5)+(\lambda-7)=0 \) \(14 \lambda=23 \Rightarrow \lambda=\frac{23}{14} \) \( \mathrm{~N}\left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right) \)…
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