Let the system of linear equations-x+2y-9z=7- | vedclass

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(D) Given equations are:$-x+2y-9z=7$ $(1)$$-x+3y-7z=9$ $(2)$$-2x+y+5z=8$ $(3)$$-3x+y+13z=\lambda$ $(4)$Subtracting $(1)$ from $(2)$:$(-x+3y-7z) - (-x+

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$7$

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(D) Given equations are:$-x+2y-9z=7$ $(1)$$-x+3y-7z=9$ $(2)$$-2x+y+5z=8$ $(3)$$-3x+y+13z=\lambda$ $(4)$Subtracting $(1)$ from $(2)$:$(-x+3y-7z) - (-x+2y-9z) = 9-7$$y+2z=2$ $(5)$Subtracting $2 	imes (1)$ from $(3)$:$(-2x+y+5z) - 2(-x+2y-9z) = 8-2(7)$$-3y+23z=-6$ $(6)$Solving $(5)$ and $(6)$ by multiplying $(5)$ by $3$ and adding to $(6)$:$3(y+2z) + (-3y+23z) = 3(2) - 6$$29z = 0 \Rightarrow z=0$Substituting $z=0$ in $(5)$:$y=2$Substituting $y=2, z=0$ in $(1)$:$-x+2(2)-9(0)=7 \Rightarrow -x+4=7 \Rightarrow x=-3$So,$(\alpha, \beta, \gamma) = (-3, 2, 0)$.Substituting these values into $(4)$ to find $\lambda$:$-3(-3) + 2 + 13(0) = \lambda \Rightarrow 9+2 = \lambda \Rightarrow \lambda = 11$.The distance of the point $(-3, 2, 0)$ from the plane $2x-2y+z-11=0$ is:$d = \frac{|2(-3) - 2(2) + 1(0) - 11|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|-6 - 4 - 11|}{\sqrt{4+4+1}} = \frac{|-21|}{3} = 7$.

Let the system of linear equations
$-x+2y-9z=7$
$-x+3y-7z=9$
$-2x+y+5z=8$
$-3x+y+13z=\lambda$
has a unique solution $x=\alpha, y=\beta, z=\gamma$. Then the distance of the point $(\alpha, \beta, \gamma)$ from the plane $2x-2y+z=\lambda$ is

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Let the system of linear equations$-x+2y-9z=7$$-x+3y-7z=9$$-2x+y+5z=8$$-3x+y+13z=\lambda$has a unique solution $x=\alpha, y=\beta, z=\gamma$. Then the distance of the point $(\alpha, \beta, \gamma)$ from the plane $2x-2y+z=\lambda$ is