JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the lines \(l_1: \frac{ x +5}{3}=\frac{ y +4}{1}=\frac{ z -\alpha}{-2}\) and \(l_2: 3 x +\) \(2 y+z-2=0=x-3 y+2 z-13\) be coplanar. If the point \(P ( a , b , c )\) on \(l_1\) is nearest to the point \(Q (-\) \(4,-3,2)\), then \(|a|+|b|+|c|\) is equal to
- A \(12\)
- B \(14\)
- C \(10\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(10\)
Step-by-step Solution
Detailed explanation
\((3 x+2 y+z-2)+\mu(x-3 y+2 z-13)=0\) \(3(3+\mu)+1 \cdot(2-3 \mu)-2(1+2 \mu)=0\) \(9-4 \mu=0\) \(\mu=\frac{9}{4}\) \(4(-15-8+\alpha-2)+9(-5+12+2 \alpha-13)=0\) \(-100+4 \alpha-54+18 \alpha=0\) \(\Rightarrow \alpha=7\)…
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