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If the length of the perpendicular drawn from the point \mathrm{P}(\mathrm{a}, 4,2)$, $\mathrm{a}>0 on the line \mathrm{\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}} is \mathrm{2 \sqrt{6}} units and \mathrm{Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)} is the image of the point \mathrm{P} in this line, then \mathrm{\mathrm{a}+\sum_{i=1}^{3} \alpha_{i}} is equal to:

Option: 1

7


Option: 2

8


Option: 3

12


Option: 4

14


Answers (1)

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\begin{aligned} &\text{DR's of P M}\mathrm{=(2 k-1-a, 3 k-1,-k-1)}\\ &\text{As } \mathrm{P M \perp line}\\ &\Rightarrow \mathrm{2(2k-1-a)+3(3k-1)-1(-k-1)=0}\\ &\Rightarrow \mathrm{ 4 k-2-2 a+9)+3(3 k-1)-1(-k-1)=0}\\ & \Rightarrow \mathrm{14 k-4-2 a=0} \\ & \Rightarrow \mathrm{7 k-2=a}\\ &\text{Also } \mathrm{ PM =2 \sqrt{6}}\\ &\Rightarrow \mathrm{P M^{2}=24} \\ &\mathrm{(-5 k+1)^{2}+(3 k-1)^{2}+(-k-1)^{2}=24} \\ &\Rightarrow \mathrm{k=1}\\ &\Rightarrow \mathrm{a=5}\\ &\therefore \mathrm{ P(5,4,2), M(1,6,0) }\\ &\text { Using mid-point formula } \\ &\mathrm{\frac{5+\alpha_{1}}{2}=1, \frac{4+\alpha_{2}}{2}=6, \frac{2+\alpha_{3}}{2}=0} \\ &\quad \mathrm{\alpha_{1}=-3, \alpha_{2}=8, \alpha_{3}=-2} \\ &\therefore\mathrm{ a+\sum \alpha} \\ &\mathrm{=5-3+8-2} \\ &\mathrm{=8 }\\ &\therefore \text { option (B) } \end{aligned}

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himanshu.meshram

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