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

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

Aicharya · Accepted Answer

Point $P(\alpha, \beta, \gamma)$ lies on plane $2 x+y+z=1 \ldots$

Form Eqs. (i) and (iv), we get $ 2 \alpha+\beta+\gamma=1 \quad \Rightarrow \alpha+\beta+\gamma=0 \quad \Rightarrow \alpha=1 $ put in Eqs. (ii) and (iii), we get $ 1+8 \beta+7 \gamma=0 $

and $\quad 9(1)+2 \beta+3 \gamma=0$

$ \begin{aligned} \text { Eq (V) }-4 \times \text { Eq. (vii) } & \\ 8 \beta+7 \gamma & =-1 \\ 8 \beta+12 \gamma & =-36 \\ & =-1 \\ \hline-5 \gamma & =35 \\ \Rightarrow \quad \gamma & =-7 \end{aligned} $ Put value of $\gamma$ in Eqs. (vi), we get $ 2 \beta+3(-7)=-9 \Rightarrow 2 \beta=12 \Rightarrow \beta=6 $ Now, $\alpha^2+\beta^2+\gamma^2$ $ =(1)^2+(6)^2+(-7)^2=1+36+49=86 $

If the point $P(\alpha, \beta, \gamma)$ lies on the plane $2 x+y+z=1$ and $[\alpha \beta…

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