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(D) Since the point $P(\alpha, \beta, \gamma)$ lies on the plane $2x + y + z = 1$,it must satisfy the equation: $2\alpha + \beta + \gamma = 1 \quad \d

Vedclass · Accepted Answer

$86$

Vedclass · Answer

(D) Since the point $P(\alpha, \beta, \gamma)$ lies on the plane $2x + y + z = 1$,it must satisfy the equation: $2\alpha + \beta + \gamma = 1 \quad \dots(i)$ Given the matrix equation: $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 9 & 1 \ 8 & 2 & 1 \ 7 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$ Performing matrix multiplication,we get: $\alpha + 8\beta + 7\gamma = 0 \quad \dots(ii)$ $9\alpha + 2\beta + 3\gamma = 0 \quad \dots(iii)$ $\alpha + \beta + \gamma = 0 \quad \dots(iv)$ Subtracting equation $(iv)$ from equation $(i)$: $(2\alpha + \beta + \gamma) - (\alpha + \beta + \gamma) = 1 - 0 \implies \alpha = 1$ Substituting $\alpha = 1$ into $(iv)$: $1 + \beta + \gamma = 0 \implies \beta + \gamma = -1 \quad \dots(v)$ Substituting $\alpha = 1$ into $(ii)$: $1 + 8\beta + 7\gamma = 0 \implies 8\beta + 7\gamma = -1 \quad \dots(vi)$ From $(v)$,$\gamma = -1 - \beta$. Substituting into $(vi)$: $8\beta + 7(-1 - \beta) = -1 \implies 8\beta - 7 - 7\beta = -1 \implies \beta = 6$ Then $\gamma = -1 - 6 = -7$. Thus,$\alpha^2 + \beta^2 + \gamma^2 = (1)^2 + (6)^2 + (-7)^2 = 1 + 36 + 49 = 86$.

If the point $P(\alpha, \beta, \gamma)$ lies on the plane $2x + y + z = 1$ and $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 9 & 1 \\ 8 & 2 & 1 \\ 7 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$,then $\alpha^2 + \beta^2 + \gamma^2 = $

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