JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let L be the line \( \frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6} \) and let S be the set of all points (a, b, c) on L, whose distance from the line \( \frac{x+1}{2}=\frac{y+1}{3}=\frac{z+9}{0} \) along the line L is 7. Then \( \sum_{(a,b,c)\in S}\ (a+b+c) \) is equal to:
- A 34
- B 28
- C 40
- D 6
Answer & Solution
Correct Answer
(A) 34
Step-by-step Solution
Detailed explanation
\(M\) is the point of intersection of \(L_1 \& L_2\) \(\Rightarrow 2 \lambda-1=2 \mu-1,3 \lambda-1=3 \mu-1,6 \lambda-3=9\) \(\Rightarrow \lambda=2=\mu\) \(\Rightarrow M (3,5,9)\) Now let point P be \((2 K-1,3 K-1,6 K-3)\) on \(L _2\) such that \(PM =7\)…
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