JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the length of the perpendicular drawn from the point \(P ( a , 4,2), a >0\) on the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}\) is \(2 \sqrt{6}\) units and \(Q \left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)\) is the image of the point \(P\) in this line, then \(a+\sum_{i=1}^{3} \alpha_{i}\) is equal to.
- A \(7\)
- B \(8\)
- C \(12\)
- D \(14\)
Answer & Solution
Correct Answer
(B) \(8\)
Step-by-step Solution
Detailed explanation
\((2 \lambda-1-a) 2+(3 \lambda-1) 3+(-\lambda-1)(-1)=0\) \(4 \lambda-2-2 a+9 \lambda-3+\lambda+1=0\) \(14 \lambda-4-2 a =0\) \(7 \lambda-2-a=0\) and, \((2 \lambda-1-a)^{2}+(3 \lambda-1)^{2}+(\lambda+1)^{2}=24\) \(5 \lambda-1)^{2}+(3 \lambda-1)^{2}+(\lambda+1)^{2}=24\)…
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