Verified Solution • 2025–26 Session moderate QJEEM24Ma59d6: JEE Main 2024 Mathematics - Class 12 JEEM24Ma59d6 Solution: Let be the image of the point in the line . Then is equal to… Let P(α,β,γ)\mathrm{P}(\alpha, \beta, \gamma)P(α,β,γ) be the image of the point Q(1,6,4)\mathrm{Q}(1,6,4)Q(1,6,4) in the line x1=y−12=z−23\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}1x=2y−1=3z−2 . Then 2α+β+γ2 \alpha+\beta+\gamma2α+β+γ is equal to ________ Step-by-Step Verified Solution amp;QR→⋅(i^+2j^+3k^)=0amp;(t−1)+(2t−5)×2+(3t−2)×3=0⇒t=1714amp;⇒R≡(1714,4814,7914)amp;⇒α+12=1714,β+62=4814,γ+42=7914amp;2α+β+γ=6814−2+9614−6+15814−4=11\begin{aligned}& \overrightarrow{Q R} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=0 \\& (t-1)+(2 t-5) \times 2+(3 t-2) \times 3=0 \Rightarrow t=\frac{17}{14} \\& \Rightarrow R \equiv\left(\frac{17}{14}, \frac{48}{14}, \frac{79}{14}\right) \\& \Rightarrow \frac{\alpha+1}{2}=\frac{17}{14}, \frac{\beta+6}{2}=\frac{48}{14}, \frac{\gamma+4}{2}=\frac{79}{14} \\& 2 \alpha+\beta+\gamma=\frac{68}{14}-2+\frac{96}{14}-6+\frac{158}{14}-4=11\end{aligned}amp;QR⋅(i^+2j^+3k^)=0amp;(t−1)+(2t−5)×2+(3t−2)×3=0⇒t=1417amp;⇒R≡(1417,1448,1479)amp;⇒2α+1=1417,2β+6=1448,2γ+4=1479amp;2α+β+γ=1468−2+1496−6+14158−4=11 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring
