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JEE Mains · Maths · STD 12 - 11. three dimension geometry

Let the equation of the plane \(P\) containing the line \(x+10=\frac{8-y}{2}=z\) be \(a x+b y+3 z=2(a+b)\) and the distance of the plane \(P\) from the point \((1,27,7)\) be c. Then \(a^2+b^2+c^2\) is equal to \(.............\).

  1. A \(354\)
  2. B \(353\)
  3. C \(355\)
  4. D \(35.5\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(355\)

Step-by-step Solution

Detailed explanation

The line \(\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z}{1}\) have a point \((-10,8,0)\) with d. r. \((1,-2,1)\) \(\because\) the plane \(a x+b y+3 z=2(a+b)\) \(\Rightarrow b =2 a\) and dot product of d.r.'s is zero \(\therefore a-2 b+3=0\) \(\therefore a =1\) and \(b =2\) Distance…
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