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QJEEM23Mae7f2: JEE Main 2023 Mathematics - Class 12 JEEM23Mae7f2 Solution: Let the lines and be coplanar. If the point on is nearest to…

Let the lines l1:x+53=y+41=z−α−2l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2} and l2:3x+2y+z−2=0=x−3y+2z−13l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13 be coplanar. If the point P(a,b,c)\mathrm{P}(a, b, c) on l1l_{1} is nearest to the point Q(−4,−3,2)\mathrm{Q}(-4,-3,2) , then ∣a∣+∣b∣+∣c∣|a|+|b|+|c| is equal to

Multiple Choice Options

  • A 12
  • B 14
  • C 10
  • D 8

Step-by-Step Verified Solution

amp;(3x+2y+z−2)+μ(x−3y+2z−13)=0amp;3(3+μ)+1⋅(2−3μ)−2(1+2μ)=0amp;9−4μ=0amp;μ=94amp;4(−15−8+α−2)+9(−5+12+2α−13)=0amp;−100+4α−54+18α=0amp;⇒α=7amp; Let P≡(3λ−5,λ−4,−2λ+7)amp; Direction ratio of PQ(3λ−1,λ−1,−2λ+5)amp; But PQ ⊥ℓ1amp;⇒3(3λ−1)+1⋅(λ−1)−2(−2λ+5)=0amp;⇒λ=1amp;P(−2,−3,5)⇒∣a∣+∣b∣+∣c∣=10\begin{aligned}& (3 \mathrm{x}+2 \mathrm{y}+\mathrm{z}-2)+\mu(\mathrm{x}-3 \mathrm{y}+2 \mathrm{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 \\\\& \text { Let } \mathrm{P} \equiv(3 \lambda-5, \lambda-4,-2 \lambda+7) \\\\& \text { Direction ratio of } \mathrm{PQ}(3 \lambda-1, \lambda-1,-2 \lambda+5) \\\\& \text { But PQ } \perp \ell_1 \\\\& \Rightarrow 3(3 \lambda-1)+1 \cdot(\lambda-1)-2(-2 \lambda+5)=0 \\\\& \Rightarrow \lambda=1 \\\\& \mathrm{P}(-2,-3,5) \Rightarrow|\mathrm{a}|+|\mathrm{b}|+|\mathrm{c}|=10\end{aligned}

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