If the point P(alpha, beta, gamma) lies on th | vedclass

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Question

(D) Given that the point $P(\alpha, \beta, \gamma)$ lies on the plane $2x+y+z=1$,we have:
$2\alpha + \beta + \gamma = 1 \quad \dots(i)$
From the mat

Vedclass · Accepted Answer

$86$

Vedclass · Answer

(D) Given that the point $P(\alpha, \beta, \gamma)$ lies on the plane $2x+y+z=1$,we have:
$2\alpha + \beta + \gamma = 1 \quad \dots(i)$
From the matrix equation $[\alpha \ \beta \ \gamma] \begin{bmatrix} 1 & 9 & 1 \ 8 & 2 & 1 \ 7 & 3 & 1 \end{bmatrix} = [0 \ 0 \ 0]$,we get the following system of linear equations:
$\alpha + 8\beta + 7\gamma = 0 \quad \dots(ii)$
$9\alpha + 2\beta + 3\gamma = 0 \quad \dots(iii)$
$\alpha + \beta + \gamma = 0 \quad \dots(iv)$
Subtracting equation $(iv)$ from equation $(i)$:
$(2\alpha + \beta + \gamma) - (\alpha + \beta + \gamma) = 1 - 0$
$\alpha = 1$
Substituting $\alpha = 1$ into equations $(ii)$ and $(iii)$:
$1 + 8\beta + 7\gamma = 0 \Rightarrow 8\beta + 7\gamma = -1 \quad \dots(v)$
$9(1) + 2\beta + 3\gamma = 0 \Rightarrow 2\beta + 3\gamma = -9 \quad \dots(vi)$
To solve for $\beta$ and $\gamma$,multiply equation $(vi)$ by $4$:
$8\beta + 12\gamma = -36 \quad \dots(vii)$
Subtracting equation $(v)$ from equation $(vii)$:
$(8\beta + 12\gamma) - (8\beta + 7\gamma) = -36 - (-1)$
$5\gamma = -35 \Rightarrow \gamma = -7$
Substituting $\gamma = -7$ into equation $(vi)$:
$2\beta + 3(-7) = -9$
$2\beta - 21 = -9 \Rightarrow 2\beta = 12 \Rightarrow \beta = 6$
Finally,calculating $\alpha^2 + \beta^2 + \gamma^2$:
$\alpha^2 + \beta^2 + \gamma^2 = (1)^2 + (6)^2 + (-7)^2 = 1 + 36 + 49 = 86$

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

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