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(B) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero $(D = 0)$.The system is:$x

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$0.5$

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(B) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero $(D = 0)$.The system is:$x - cy - cz = 0$$cx - y + cz = 0$$cx + cy - z = 0$The determinant $D$ is given by:$D = \begin{vmatrix} 1 & -c & -c \ c & -1 & c \ c & c & -1 \end{vmatrix} = 0$Expanding the determinant along the first row:$1((-1)(-1) - (c)(c)) - (-c)((c)(-1) - (c)(c)) + (-c)((c)(c) - (-1)(c)) = 0$$1(1 - c^2) + c(-c - c^2) - c(c^2 + c) = 0$$1 - c^2 - c^2 - c^3 - c^3 - c^2 = 0$$-2c^3 - 3c^2 + 1 = 0$$2c^3 + 3c^2 - 1 = 0$Factoring the cubic equation:$(c + 1)^2(2c - 1) = 0$The roots are $c = -1$ (with multiplicity $2$) and $c = \frac{1}{2}$.Therefore,the greatest value of $c$ is $\frac{1}{2}$ or $0.5$.

The greatest value of $c \in R$ for which the system of linear equations $x - cy - cz = 0$,$cx - y + cz = 0$,$cx + cy - z = 0$ has a non-trivial solution,is

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