JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(P\) be the plane containing the straight line \(\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}\) and perpendicular to the plane containing the straight lines \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\) and \(\frac{x}{3}=\frac{y}{7}=\frac{z}{8}\). If \(d\) is the distance of \(P\) from the point \((2,-5,11)\), then \(d ^{2}\) is equal to.
- A \(\frac{147}{2}\)
- B \(96\)
- C \(\frac{32}{3}\)
- D \(54\)
Answer & Solution
Correct Answer
(C) \(\frac{32}{3}\)
Step-by-step Solution
Detailed explanation
\(a(x-3)+b(y+4)+c(z-7)=0\) \(P: \quad 9 a-b-5 c=0\) \(\quad-11 a-b+5 c=0\) After solving \(DR\) 's \(\propto(1,-1,2)\) Equation of plane \(x-y+2 z=21\) \(d=\frac{8}{\sqrt{6}}\) \(d^{2}=\frac{32}{3}\)
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