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(C) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.The determinant is

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$-1$

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(C) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.The determinant is given by:$\Delta = \left|\begin{array}{ccc}a^3 & (a+1)^3 & (a+2)^3 \ a & a+1 & a+2 \ 1 & 1 & 1\end{array}ight| = 0$Apply column operations: $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_2$:$\Delta = \left|\begin{array}{ccc}a^3 & (a+1)^3 - a^3 & (a+2)^3 - (a+1)^3 \ a & 1 & 1 \ 1 & 0 & 0\end{array}ight| = 0$Expanding along the third row:$1 \cdot \left[((a+1)^3 - a^3) - ((a+2)^3 - (a+1)^3)ight] = 0$$(a+1)^3 - a^3 - (a+2)^3 + (a+1)^3 = 0$$2(a+1)^3 - a^3 - (a+2)^3 = 0$$2(a^3 + 3a^2 + 3a + 1) - a^3 - (a^3 + 6a^2 + 12a + 8) = 0$$2a^3 + 6a^2 + 6a + 2 - a^3 - a^3 - 6a^2 - 12a - 8 = 0$$-6a - 6 = 0$$-6a = 6 \implies a = -1$Thus,the value of $a$ is $-1$.

The value of $a$ for which the system of equations $a^3x + (a + 1)^3y + (a + 2)^3z = 0$,$ax + (a + 1)y + (a + 2)z = 0$,and $x + y + z = 0$ has a non-zero solution is:

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