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(D) The system of equations has a unique solution if the determinant $\Delta 
eq 0$.$\Delta = \begin{vmatrix} a & 2a & -3a \ 2a+1 & 2a+3 & a+1 \ 3a

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$S_1 = R - \{0\}$ and $S_2 = \Phi$

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(D) The system of equations has a unique solution if the determinant $\Delta 
eq 0$.$\Delta = \begin{vmatrix} a & 2a & -3a \ 2a+1 & 2a+3 & a+1 \ 3a+5 & a+5 & a+2 \end{vmatrix}$Taking $a$ common from the first column:$\Delta = a \begin{vmatrix} 1 & 2a & -3a \ 2a+1 & 2a+3 & a+1 \ 3a+5 & a+5 & a+2 \end{vmatrix}$Performing row operations $R_2 	o R_2 - (2a+1)R_1$ and $R_3 	o R_3 - (3a+5)R_1$:$\Delta = a \begin{vmatrix} 1 & 2a & -3a \ 0 & 2a+3 - 2a(2a+1) & a+1 + 3a(2a+1) \ 0 & a+5 - 2a(3a+5) & a+2 + 3a(3a+5) \end{vmatrix}$Simplifying the entries:$\Delta = a \begin{vmatrix} 1 & 2a & -3a \ 0 & -4a^2+3 & 6a^2+4a+1 \ 0 & -6a^2-9a+5 & 9a^2+16a+2 \end{vmatrix}$Calculating the determinant of the $2 	imes 2$ matrix:$(-4a^2+3)(9a^2+16a+2) - (6a^2+4a+1)(-6a^2-9a+5) = 0$After expansion,we find $\Delta = 0$ only when $a = 0$. Since $a \in R - \{0\}$,$\Delta$ is never $0$ for any $a$ in the set.Thus,the system always has a unique solution for all $a \in R - \{0\}$.Therefore,$S_1 = R - \{0\}$ and $S_2 = \Phi$.

Let $S_1$ and $S_2$ be respectively the sets of all $a \in R - \{0\}$ for which the system of linear equations
$a x + 2 a y - 3 a z = 1$
$(2 a + 1) x + (2 a + 3) y + (a + 1) z = 2$
$(3 a + 5) x + (a + 5) y + (a + 2) z = 3$
has a unique solution and infinitely many solutions,respectively. Then:

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Let $S_1$ and $S_2$ be respectively the sets of all $a \in R - \{0\}$ for which the system of linear equations$a x + 2 a y - 3 a z = 1$$(2 a + 1) x + (2 a + 3) y + (a + 1) z = 2$$(3 a + 5) x + (a + 5) y + (a + 2) z = 3$has a unique solution and infinitely many solutions,respectively. Then: