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

If the equation of a plane \(P ,\) passing through the intesection of the planes, \(x+4 y-z+7=0\) and \(3 x+y+5 z=8\) is \(ax +b y+6 z=15\) for some \(a, b \in R,\) then the distance of the point \((3,2,-1)\) from the plane \(P\) is

  1. A \(3\)
  2. B \(7\)
  3. C \(21\)
  4. D \(63\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(3\)

Step-by-step Solution

Detailed explanation

\(D _{1}=\left|\begin{array}{ccc}-7 & 4 & -1 \\ 8 & 1 & 5 \\ 15 & b & 6\end{array}\right|=0 \Rightarrow b =-3\) \(D=\left|\begin{array}{ccc}1 & 4 & -1 \\ 3 & 1 & 5 \\ a & b & 6\end{array}\right|=0 \Rightarrow 21 a-8 b-66=0 \ldots\) \(P: 2 x-3 y+6 z=15\) so required distance…