If the system of equations 2x + 9y + 5z = 8,2 | vedclass

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(D) Given the system of equations $2x + 9y + 5z = 8$,$2x + 3y - z = -4$,and $x - 2z = -5$ has an infinite number of solutions given by $x = -5 + at$,$

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$2, -1, 1$

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(D) Given the system of equations $2x + 9y + 5z = 8$,$2x + 3y - z = -4$,and $x - 2z = -5$ has an infinite number of solutions given by $x = -5 + at$,$y = 2 + bt$,$z = ct$,where $t \in R$. Substituting these into the equations,we get: $2(-5 + at) + 9(2 + bt) + 5(ct) = 8$ $2(-5 + at) + 3(2 + bt) - (ct) = -4$ $(-5 + at) - 2(ct) = -5$ Simplifying these: $-10 + 2at + 18 + 9bt + 5ct = 8 \Rightarrow 2at + 9bt + 5ct = 0$ $-10 + 2at + 6 + 3bt - ct = -4 \Rightarrow 2at + 3bt - ct = 0$ $-5 + at - 2ct = -5 \Rightarrow at - 2ct = 0$ Dividing by $t$ (assuming $t 
eq 0$): $2a + 9b + 5c = 0$ $2a + 3b - c = 0$ $a - 2c = 0 \Rightarrow a = 2c$ Substituting $a = 2c$ into $2a + 3b - c = 0$: $2(2c) + 3b - c = 0 \Rightarrow 4c + 3b - c = 0 \Rightarrow 3b = -3c \Rightarrow b = -c$ Thus,$a : b : c = 2c : -c : c = 2 : -1 : 1$. Therefore,$a = 2$,$b = -1$,$c = 1$.

If the system of equations $2x + 9y + 5z = 8$,$2x + 3y - z = -4$,$x - 2z = -5$ has an infinite number of solutions $x = -5 + at$,$y = 2 + bt$,$z = ct$,$t \in R$,then $a$,$b$,$c$ respectively are

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