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(D) Given the matrix equation $(\alpha \ \beta \ \gamma)\begin{bmatrix} 2 & 10 & 8 \ 9 & 3 & 8 \ 8 & 4 & 8 \end{bmatrix} = (0 \ 0 \ 0)$,we get the f

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$11$

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(D) Given the matrix equation $(\alpha \ \beta \ \gamma)\begin{bmatrix} 2 & 10 & 8 \ 9 & 3 & 8 \ 8 & 4 & 8 \end{bmatrix} = (0 \ 0 \ 0)$,we get the following system of linear equations:$2\alpha + 9\beta + 8\gamma = 0 \quad (1)$$10\alpha + 3\beta + 4\gamma = 0 \quad (2)$$8\alpha + 8\beta + 8\gamma = 0 \quad (3)$From $(3)$,we have $\alpha + \beta + \gamma = 0$,which implies $\gamma = -\alpha - \beta$.Substitute $\gamma$ into $(1)$:$2\alpha + 9\beta + 8(-\alpha - \beta) = 0 \implies -6\alpha + \beta = 0 \implies \beta = 6\alpha$.Now,find $\gamma$ in terms of $\alpha$:$\gamma = -\alpha - 6\alpha = -7\alpha$.The point $P(\alpha, 6\alpha, -7\alpha)$ lies on the plane $2x + 4y + 3z = 5$:$2(\alpha) + 4(6\alpha) + 3(-7\alpha) = 5$$2\alpha + 24\alpha - 21\alpha = 5$$5\alpha = 5 \implies \alpha = 1$.Thus,$\alpha = 1, \beta = 6, \gamma = -7$.We need to calculate $6\alpha + 9\beta + 7\gamma$:$6(1) + 9(6) + 7(-7) = 6 + 54 - 49 = 11$.

If a point $P(\alpha, \beta, \gamma)$ satisfying $(\alpha \ \beta \ \gamma)\begin{bmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{bmatrix} = (0 \ 0 \ 0)$ lies on the plane $2x + 4y + 3z = 5$,then $6\alpha + 9\beta + 7\gamma$ is equal to:

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