If a point P(α, β, γ) satisfying ( cases α & β & γ cases ) ( cases...

Given,

\left(\begin{cases}\alpha & \beta & \gamma\end{cases}\right)\left(\begin{cases}2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8\end{cases}\right) = \left(\begin{cases}0 & 0 & 0\end{cases}\right)

⇒ \left(\begin{cases}2\alpha + 9\beta + 8\gamma & 10\alpha + 3\beta + 4\gamma & 8\alpha + 8\beta + 8\gamma\end{cases}\right) = \left(\begin{cases}0 & 0 & 0\end{cases}\right)

Now on comparing both side we get,

$2\alpha + 2\beta + 8\gamma = 0 \ldots \left(1\right)$

$10\alpha + 3\beta + 4\gamma = 0 \ldots \left(2\right)$

$8\alpha + 8\beta + 8\gamma = 0 \ldots \left(3\right)$

Now from \left(1\right)$and \left(3\right)$ by cross multiplication we get,

$\dfrac{\alpha}{1} = \dfrac{\beta}{6} = \dfrac{\gamma}{- 7} = k ⇒ \alpha = k, \beta = 6k, \gamma = - 7k$

Point $P\left(\alpha, \beta, \gamma\right)$ lie on the plane $2x + 4y + 3z = 5 ⇒ 2\alpha + 4\beta + 3\gamma = 5$

$⇒ 2k + 24k - 21k = 5$

$⇒ k = 1$

Hence, $6\alpha + 9\beta + 7\gamma = 6k + 54k - 49k = 11k = 11$.