Correct option is (3) 108
\(\mathrm L: \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}=\lambda, \mathrm p(3,4,9)\)
\(\mathrm{M}(3 \lambda+1,2 \lambda-1, \lambda+2)\)
\( \mathrm{\overrightarrow{PM}}=(3 \lambda-2,2 \lambda-5, \lambda-7)\)
DR's of \(\mathrm L:(3,2,1)\)
\(\mathrm{\overrightarrow{PM}} \perp\text{ line L}\)
\(\text {So, } 3(3 \lambda-2)+2(2 \lambda-5)+1(\lambda-7)=0\)
\(9 \lambda-6+4 \lambda-10+\lambda-7=0\)
\(14 \lambda=23 \Rightarrow \lambda=\frac{23}{14}\)
\(\mathrm{M}\left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right)\)
Now, As \(\mathrm{M}\) is mid-point of \(\mathrm{pp}^{\prime}\)
\(\therefore\) co-ordinates of \(\mathrm{p}^{\prime}\) are
\(\frac{\alpha+3}{2}=\frac{83}{14} \Rightarrow \alpha=\frac{62}{7}\)
\(\frac{\beta+4}{2}=\frac{32}{14} \Rightarrow \beta=\frac{4}{7} \)
\(\frac{\gamma+9}{2}=\frac{51}{14} \Rightarrow \gamma=\frac{-12}{7}\)
So, \(14(\alpha+\beta+\gamma)=14 \times \frac{54}{7} \Rightarrow 108\)
