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AP EAMCET · Maths · Three Dimensional Geometry

Let \(\pi\) be the plane passing through the point \((3,-3,1)\) and perpendicular to the line joining the points \((3,4,-1)\) and \((2,-1,5)\). If the equation of the plane containing the points \((3,4,-1),(-1,2,5)\) and perpendicular to the plane \(\pi\) is \(a x+y+c z-d=0\), then \(3(a+c)=\)

  1. A \(-d\)
  2. B \(2 d\)
  3. C \(d\)
  4. D \(-2 d\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(d\)

Step-by-step Solution

Detailed explanation

Plane \(\pi\) passes through \((3,-3,1)\) and perpendicular to the line joining the points \((3,4,-1)\) and \((2,-1,5)\). \(\therefore \mathrm{DR}\) 's of normal to the plane are \((3-2),(4,+1),(-1-5)\) \(\equiv(1,5,-6)\) Equation of plane \(\pi\) is given by \(x+5 y-6 z+d=0\)…
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