JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the plane \(P\) pass through the intersection of the planes \(2 x+3 y-z=2\) and \(x+2 y+3 z=6\), and be perpendicular to the plane \(2 x+y-z+1=0\). If \(d\) is the distance of \(P\) from the point \((-7,1,1)\), then \(d ^2\) is equal to :
- A \(\frac{250}{83}\)
- B \(\frac{15}{53}\)
- C \(\frac{25}{83}\)
- D \(\frac{250}{82}\)
Answer & Solution
Correct Answer
(A) \(\frac{250}{83}\)
Step-by-step Solution
Detailed explanation
\(P \equiv P _1+\lambda P _2=0\) \((2+\lambda) x +(3+2 \lambda) y+(3 \lambda-1) z-2-6 \lambda=0\) Plane \(P\) is perpendicular to \(P_3 \therefore \vec{n} \cdot \vec{n}_3=0\) \(2(\lambda+2)+(2 \lambda+3)-(3 \lambda-1)=0\) \(\lambda=-8\) \(P \equiv-6 x-13 y-25 z+46=0\)…
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