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(B) The system of equations has no common point if the determinant of the coefficient matrix is zero and the system is inconsistent.The coefficient ma

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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 is zero and the system is inconsistent.The coefficient matrix $D$ is given by:$D = \begin{vmatrix} a+1 & -1 & -1 \ 1 & -(a+1) & -1 \ 1 & -1 & -(a+1) \end{vmatrix}$Applying $R_1 	o R_1 + R_2 + R_3$,we get:$D = \begin{vmatrix} a+1 & -(a+2) & -(a+2) \ 1 & -(a+1) & -1 \ 1 & -1 & -(a+1) \end{vmatrix} = (a-1) \begin{vmatrix} 1 & -1 & -1 \ 1 & -(a+1) & -1 \ 1 & -1 & -(a+1) \end{vmatrix}$Solving the determinant,we find $D = -(a-1)^2(a+3) = 0$,so $a = 1$ or $a = -3$.If $a = 1$,the planes are $2x-y-z=1, x-2y-z=1, x-y-2z=1$,which intersect at a point.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,the planes have no common point when $a = -3$.

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