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

Let \(\alpha x+\beta y+y z=1\) be the equation of a plane passing through the point \((3,-2,5)\) and perpendicular to the line joining the points \((1, 2, 3)\) and \((-2,3,5)\). Then the value of \(\alpha \beta \gamma\) is equal to \(..........\).

  1. A \(5\)
  2. B \(6\)
  3. C \(4\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(6\)

Step-by-step Solution

Detailed explanation

Given Equation is not equation of plane as \(yz\) is present. If we consider \(y\) is \(\gamma\) then answer would be 6. Normal vector of plane \(=3 \hat{i}-\hat{j}-2 \hat{k}\) Plane : \(3 x-y-2 z+\lambda=0\) Point \((3,-2,5)\) satisfies the plane \(\lambda=-1\) \(3 x-y-2 z=1\)…
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