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(B) The system of equations has no common point if the determinant of the coefficient matrix $D = 0$ and the system is inconsistent.The determinant $D

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$-3$

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(B) The system of equations has no common point if the determinant of the coefficient matrix $D = 0$ and the system is inconsistent.The determinant $D$ is given by:$D = \begin{vmatrix} a+1 & -1 & -1 \ 1 & -(a+1) & -1 \ 1 & -1 & -(a+1) \end{vmatrix}$Applying $R_1 	o R_1 + R_2 + R_3$:$D = \begin{vmatrix} a+1 & -(a+2) & -(a+2) \ 1 & -(a+1) & -1 \ 1 & -1 & -(a+1) \end{vmatrix} = (a-1) \begin{vmatrix} 1 & -1 & -1 \ 1 & -(a+1) & -1 \ 1 & -1 & -(a+1) \end{vmatrix}$Alternatively,calculating directly:$D = (a+1)[(a+1)^2 - 1] + 1[-(a+1) + 1] - 1[-1 + (a+1)]$$D = (a+1)(a^2 + 2a) - a - a = a(a+1)(a+2) - 2a = a(a^2 + 3a + 2 - 2) = a^2(a+3)$For no common point,we require $D = 0$,which gives $a = 0$ or $a = -3$.If $a = 0$,the planes are $x - y - z = 0$,$x - y - z = 0$,$x - y - z = 0$,which are identical and have infinite common points.If $a = -3$,the planes are $-2x - y - z = -3$,$x + 2y - z = -3$,$x - y + 2z = -3$. Adding these gives $0 = -9$,which is impossible. Thus,there is no common point for $a = -3$.

The value of $a$ for which the three given planes
$P_1 : (a + 1)x - y - z = a$
$P_2 : x - (a + 1)y - z = a$
$P_3 : x - y - (a + 1)z = a$
have no point in common,is

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The value of $a$ for which the three given planes$P_1 : (a + 1)x - y - z = a$$P_2 : x - (a + 1)y - z = a$$P_3 : x - y - (a + 1)z = a$have no point in common,is