JEE Math Practice Question
If a point $\mathrm{P}(\alpha, \beta, \gamma)$ satisfying $$\left( {\begin{matrix} \alpha & \beta & \gamma \\ \end{matrix} } \right)\left( {\begin{matrix} 2 & {10} & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \\ \end{matrix} } \right) = \left( {\begin{matrix} 0 & 0 & 0 \\ \end{matrix} } \right)$$ lies on the plane $2 x+4 y+3 z=5$, then $6 \alpha+9 \beta+7 \gamma$ is equal to :
- A.$\frac{11}{5}$
- B.11
- C.$-1$
- D.$\frac{5}{4}$
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