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(B) Given the system of linear equations:$x+y+z=a$ ... $(i)$$x-y+bz=2$ ... $(ii)$$2x+3y-z=1$ ... $(iii)$For the system to have infinitely many solutio

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$3$

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(B) Given the system of linear equations:$x+y+z=a$ ... $(i)$$x-y+bz=2$ ... $(ii)$$2x+3y-z=1$ ... $(iii)$For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 1 & 1 & 1 \ 1 & -1 & b \ 2 & 3 & -1 \end{vmatrix} = 0$$1(1-3b) - 1(-1-2b) + 1(3+2) = 0$$1-3b + 1+2b + 5 = 0$$7-b = 0 \Rightarrow b=7$Now,substitute $b=7$ into the equations:$(i) + (ii) \Rightarrow 2x + (1+7)z = a+2 \Rightarrow 2x+8z = a+2 \Rightarrow x+4z = \frac{a+2}{2}$ ... $(iv)$Multiply $(i)$ by $3$ and subtract $(iii)$:$3(x+y+z) - (2x+3y-z) = 3a - 1$$3x+3y+3z - 2x-3y+z = 3a-1$$x+4z = 3a-1$ ... $(v)$For infinitely many solutions,$(iv)$ and $(v)$ must be identical:$\frac{a+2}{2} = 3a-1$$a+2 = 6a-2$$5a = 4$Finally,$b-5a = 7-4 = 3$.

If the system of simultaneous linear equations $x+y+z=a$,$x-y+bz=2$,and $2x+3y-z=1$ has infinitely many solutions,then $b-5a=$

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