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(C) For a system of homogeneous linear equations to have non-trivial solutions, the determinant of the coefficient matrix $A$ must be zero, i.e.,$|A|

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$1$

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(C) For a system of homogeneous linear equations to have non-trivial solutions, the determinant of the coefficient matrix $A$ must be zero, i.e.,$|A| = 0$.Given the matrix $A = \begin{bmatrix} t & t+1 & t-1 \ t+1 & t & t+2 \ t-1 & t+2 & t \end{bmatrix}$.We set the determinant $|A| = 0$:$|A| = \begin{vmatrix} t & t+1 & t-1 \ t+1 & t & t+2 \ t-1 & t+2 & t \end{vmatrix} = 0$.Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$|A| = \begin{vmatrix} t & t+1 & t-1 \ 1 & -1 & 3 \ -1 & 1 & 1 \end{vmatrix} = 0$.Adding $R_2$ to $R_3$ $(R_3 ightarrow R_3 + R_2)$:$|A| = \begin{vmatrix} t & t+1 & t-1 \ 1 & -1 & 3 \ 0 & 0 & 4 \end{vmatrix} = 0$.Expanding along the third row:$4 	imes \begin{vmatrix} t & t+1 \ 1 & -1 \end{vmatrix} = 0$.$4 	imes (-t - (t+1)) = 0$.$4 	imes (-2t - 1) = 0$.$-8t - 4 = 0 \Rightarrow t = -\frac{1}{2}$.Since there is only one real value of $t$ $(t = -\frac{1}{2})$, the number of real values is $1$.

The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is

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The number of real values of $t$ such that the system of homogeneous equations$\begin{aligned}t x+(t+1) y+(t-1) z &=0 \\(t+1) x+t y+(t+2) z &=0 \\(t-1) x+(t+2) y+t z &=0\end{aligned}$has non-trivial solutions is