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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. $\left|\begin{arr

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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. $\left|\begin{array}{ccc} a^3 & (a+1)^3 & (a+2)^3 \ a & (a+1) & (a+2) \ 1 & 1 & 1 \end{array}ight| = 0$ Interchanging rows to simplify: $-\left|\begin{array}{ccc} 1 & 1 & 1 \ a & (a+1) & (a+2) \ a^3 & (a+1)^3 & (a+2)^3 \end{array}ight| = 0$ Using the property of the Vandermonde-like determinant $\left|\begin{array}{ccc} 1 & 1 & 1 \ x & y & z \ x^3 & y^3 & z^3 \end{array}ight| = (x-y)(y-z)(z-x)(x+y+z)$,where $x=a, y=a+1, z=a+2$: $-(a-(a+1))((a+1)-(a+2))((a+2)-a)(a+(a+1)+(a+2)) = 0$ $-(-1)(-1)(2)(3a+3) = 0$ $-2(3a+3) = 0$ $3a+3 = 0 \Rightarrow a = -1$.

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

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