JEE Mains · Maths · STD 12 - 11. three dimension geometry
If a point \(P (\alpha, \beta, \gamma)\) satisfying \((\alpha \ \beta \ \gamma)\left(\begin{array}{ccc}2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8\end{array}\right)=\left(\begin{array}{llll}0 & 0 & 0\end{array}\right)\) lies on the plane \(2 x+4 y+3 z=5\), then \(6 \alpha+9 \beta+7 \gamma\) is equal to:
- A \(-1\)
- B \(\frac{11}{5}\)
- C \(\frac{5}{4}\)
- D \(11\)
Answer & Solution
Correct Answer
(D) \(11\)
Step-by-step Solution
Detailed explanation
\(2 \alpha+4 \beta+3 \gamma=5......(1)\) \(2 \alpha+9 \beta+8 \gamma=0......(2)\) \(10 \alpha+3 \beta+4 \gamma=0......(3)\) \(8 \alpha+8 \beta+8 \gamma=0......(4)\) Subtract \((4)\) from \((2)\) \(-6 \alpha+\beta=0\) \(\beta=6 \alpha\) From equation \((4)\)…
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