If a point $\mathrm{P}(\alpha, \beta, \gamma)$ satisfying

$$\left( {\matrix{ \alpha & \beta & \gamma \cr } } \right)\left( {\matrix{ 2 & {10} & 8 \cr 9 & 3 & 8 \cr 8 & 4 & 8 \cr } } \right) = \left( {\matrix{ 0 & 0 & 0 \cr } } \right)$$

lies on the plane $2 x+4 y+3 z=5$, then $6 \alpha+9 \beta+7 \gamma$ is equal to :

Point $\mathrm{P}(\alpha, \beta, \gamma)$ lies on the plane $2 x+4 y+3 z=5$, <br/><br/>$\therefore$ $2 \alpha+4 \beta+3 \gamma=5$ ........(1) <br/><br/>Given, $$\left( {\matrix{ \alpha & \beta & \gamma \cr } } \right)\left( {\matrix{ 2 & {10} & 8 \cr 9 & 3 & 8 \cr 8 & 4 & 8 \cr } } \right) = \left( {\matrix{ 0 & 0 & 0 \cr } } \right)$$ <br/><br/>$\therefore$ $2 \alpha+9 \beta+8 \gamma=0$ .......(2) <br/><br/>and $10 \alpha+3 \beta+4 \gamma=0$ ........(3) <br/><br/>and $8 \alpha+8 \beta+8 \gamma=0$ ..........(4) <br/><br/>Subtract (4) from (2) <br/><br/>$-6 \alpha+\beta=0$ $\beta=6 \alpha$ <br/><br/>From equation (4) <br/><br/>$8 \alpha+48 \alpha+8 \gamma=0$ <br/><br/>$\gamma=-7 \alpha$ <br/><br/>From equation (1) <br/><br/>$2 \alpha+24 \alpha-21 \alpha=5$ <br/><br/>$5 \alpha=5$ <br/><br/>$\alpha=1$ <br/><br/>$\beta=+6, \quad \gamma=-7$ <br/><br/>$\therefore 6 \alpha+9 \beta+7 \gamma$ <br/><br/>$=6+54-49$ <br/><br/>$=11$