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Let $A=\left(\begin{array}{c}0 \\ -6 \\ 8\end{array}\right), B=\left(\begin{array}{ccc}3 & 5 & -7 \\ 0 & -1 & 8 \\ 6 & -1 & 0\end{array}\right)$ and $X=\left(\begin{array}{l}x \\ y \\ z\end{array}\right)$. If $\mathrm{D}=[\alpha \beta \gamma]^{\mathrm{T}}$ is the solution of $\mathrm{X}^{\mathrm{T}} \mathrm{B}^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}$, then $\mathrm{D}^{\mathrm{T}} \mathrm{A}=$
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4
$\begin{aligned} & \text { } \because X^T B^T=A^T \\ & \Rightarrow\left(X^T B^T\right)=A^T \Rightarrow B X=A \\ & |B|=\left|\begin{array}{ccc}3 & 5 & -7 \\ 0 & -1 & 8 \\ 6 & -1 & 0\end{array}\right|=222, \quad B^{-1}=\frac{1}{222}\left|\begin{array}{lll}8 & 7 & 33 \\ 48 & 42 & -24 \\ 6 & 33 & -3\end{array}\right| \\ & X=B^{-1} A=\frac{1}{222}\left[\begin{array}{lll}8 & 7 & 33 \\ 48 & 42 & -24 \\ 6 & 33 & -3\end{array}\right]\left[\begin{array}{l}0 \\ -6 \\ 8\end{array}\right] \\ & {\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{222}\left[\begin{array}{c}222 \\ -444 \\ -222\end{array}\right]=\left[\begin{array}{c}1 \\ -2 \\ -1\end{array}\right]=D=\left[\begin{array}{l}\alpha \\ \beta \\ \gamma\end{array}\right]} \\ & \text { Now, } D^T A=\left[\begin{array}{lll}1 & -2 & -1\end{array}\right]\left[\begin{array}{c}0 \\ -6 \\ 8\end{array}\right]=[0+12-8]=4 \\ & \end{aligned}$
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