Number of values of m for which the lines x + | vedclass

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(A) For the lines to be concurrent,the determinant of the coefficients must be zero:$\begin{vmatrix} 1 & 1 & -1 \ m - 1 & m^2 - 7 & -5 \ m - 2 & 2m

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(A) For the lines to be concurrent,the determinant of the coefficients must be zero:$\begin{vmatrix} 1 & 1 & -1 \ m - 1 & m^2 - 7 & -5 \ m - 2 & 2m - 5 & 0 \end{vmatrix} = 0$Expanding along the third row:$-(m - 2) \begin{vmatrix} 1 & -1 \ m^2 - 7 & -5 \end{vmatrix} + (2m - 5) \begin{vmatrix} 1 & -1 \ m - 1 & -5 \end{vmatrix} = 0$$-(m - 2)(-5 + m^2 - 7) + (2m - 5)(-5 + m - 1) = 0$$-(m - 2)(m^2 - 12) + (2m - 5)(m - 6) = 0$$-(m^3 - 6m^2 - 12m + 24) + (2m^2 - 12m - 5m + 30) = 0$$-m^3 + 6m^2 + 12m - 24 + 2m^2 - 17m + 30 = 0$$-m^3 + 8m^2 - 5m + 6 = 0 \implies m^3 - 8m^2 + 5m - 6 = 0$Testing for roots,we find that for $m=3$,the lines are not concurrent because the lines become $x+y-1=0$,$2x+2y-5=0$,and $x+y=0$. The lines $x+y-1=0$ and $x+y=0$ are parallel,so they never intersect. Thus,there are no values of $m$ for which the lines are concurrent.

Number of values of $m$ for which the lines $x + y - 1 = 0$,$(m - 1)x + (m^2 - 7)y - 5 = 0$ and $(m - 2)x + (2m - 5)y = 0$ are concurrent,are

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