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Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations
$x+2 y+3 z=\alpha$
$4 x+5 y+6 z=\beta$
$7 x+8 y+9 z=\gamma-$
is consistent. Let $| M |$ represent the determinant of the matrix
$M=\left[\begin{array}{ccc}\alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$
Let $P$ be the plane containing all those $(\alpha, \beta, \gamma)$ for which the above system of linear equations is consistent, and $D$ be the square of the distance of the point $(0,1,0)$ from the plane $P$.
($1$) The value of $| M |$ is
($2$) The value of $D$ is
$1,1.5$
$1,1.6$
$1,1.7$
$1,1.8$
Solution
$7 x+8 y+9 z-(\gamma-1)=A(4 x+5 y+6 z-\beta)+B(x+2 y+3 z-\alpha)$
$x: 7=4 A+B$
$y: 8=5 A+2 B$
$A=2, B=-1$
$\text { const. term : }-(\gamma-1)=-A \beta-\alpha B \Rightarrow-(\gamma-1) \equiv 2 \beta+\alpha \alpha-2 \beta+\gamma=1$
$M=\left(\begin{array}{ccc}\alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1\end{array}\right)=\alpha-2 \beta+\gamma=1$
Plane $P : x -2 y + z =1$
$\text { Perpendicular distance }=\left|\frac{3}{\sqrt{6}}\right|= P \Rightarrow D = P ^2=\frac{9}{6}=1.5$
