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(C) For a system of homogeneous linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. $\Delta = \be

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$2 t^2+3 t+1=0$

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(C) For a system of homogeneous linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. $\Delta = \begin{vmatrix} t & t+1 & t-1 \ t+1 & t & t+2 \ t-1 & t+2 & t \end{vmatrix} = 0$ Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$: $\begin{vmatrix} t & t+1 & t-1 \ 1 & -1 & 3 \ -1 & 1 & 1 \end{vmatrix} = 0$ Expanding along the first row: $t(-1 - 3) - (t+1)(1 + 3) + (t-1)(1 - 1) = 0$ $-4t - 4(t+1) + 0 = 0$ $-4t - 4t - 4 = 0$ $-8t = 4 \Rightarrow t = -\frac{1}{2}$ Checking the options for $t = -\frac{1}{2}$: For option $(C)$, $2t^2 + 3t + 1 = 2(-\frac{1}{2})^2 + 3(-\frac{1}{2}) + 1 = 2(\frac{1}{4}) - \frac{3}{2} + 1 = \frac{1}{2} - \frac{3}{2} + 1 = 0$. Thus, $t = -\frac{1}{2}$ is a root of $2t^2 + 3t + 1 = 0$.

If the system of homogeneous equations $\begin{aligned} & t x+(t+1) y+(t-1) z=0 \\ & (t+1) x+t y+(t+2) z=0 \\ & (t-1) x+(t+2) y+t z=0\end{aligned}$ in $x, y, z$ has a non-trivial solution, then $t$ is a root of the equation

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