The value of 'a' for which the three given pl | vedclass

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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.$D = \begin{vmatri

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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.$D = \begin{vmatrix} a+1 & -1 & -1 \ 1 & -(a+1) & -1 \ 1 & -1 & -(a+1) \end{vmatrix} = (a+1)((a+1)^2 - 1) + 1(-(a+1) + 1) - 1(-1 + (a+1))$$D = (a+1)(a^2 + 2a) - a - a = (a+1)a(a+2) - 2a = a(a^2 + 3a + 2 - 2) = a^2(a+3)$.Setting $D = 0$,we get $a = 0$ or $a = -3$.For $a = 0$,the planes are $x-y-z=0$,$x-y-z=0$,$x-y-z=0$,which are identical and have infinite common points.For $a = -3$,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 system has 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