$(a)\;0\qquad(b)\;3\qquad(c)\;5\qquad(d)\;6$

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- Three lines $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$ and $a_3x+b_3y+c_3=0$ are concurrent if $\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0$

Given that the lines $ax+y-1=0,\:\:x+ay=0\:\:and\:\:x+y+a=0$

are concurrent.

$\therefore\:\begin{vmatrix}a&1&-1\\1&a&0\\1&1&a\end{vmatrix}=0$

$\Rightarrow\:a(a^2-0)-1(a-0)-1(1-a)=0$

$\Rightarrow a^3-a+a-1=0$

$\Rightarrow a^3-1=0$

Hence (a) is the correct answer.

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