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 \(..........\).
- A \(5\)
- B \(6\)
- C \(4\)
- D \(9\)
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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