Let S _ 1 and S _ 2 be respectively the sets of all a ∈ R - \ 0\...

Given,

${S}_{1}$ and ${S}_{2}$ be respectively the sets of all $a \in R - \left\{0\right\}$ for which the system of linear equations

$ax + 2ay - 3az = 1 . . . . . . . . . \left(1\right)$

\left(2a + 1\right)x + \left(2a + 3\right)y + \left(a + 1\right)z = 2 . . . . \left(2\right)$</p><p>\left(3a + 5\right)x + \left(a + 5\right)y + \left(a + 2\right)z = 3 . . . . . \left(3\right)$

Now from above equations finding \Delta = \left|\begin{cases}a & 2a & - 3a \\ 2a + 1 & 2a + 3 & a + 1 \\ 3a + 5 & a + 5 & a + 2\end{cases}\right|

$= a\left(15{a}^{2} + 31a + 36\right) = 0 ⇒ a = 0$

As $15{a}^{2} + 31a + 36$ cannot be zero as ${31}^{2} - 4 \times 15 \times 36 < 0$

So, $\Delta \ne 0$ for all $a \in R - \left\{0\right\}$

Hence, ${S}_{1} = R - \left\{0\right\} \text{ and } {S}_{2} = ϕ$