To solve the system $\left\{\begin{array}{l} 3x - 4z = 0 \\ 3y + 2z = -3 \\ 2x + 3y = -5 \end{array}\right.$ by elimination, notice that each given algebraic equation represents a relationship between only two of the three variables. First, select a pair of equations and manipulate them to eliminate a shared variable. For example, multiplying the second equation by $2$ yields $6y + 4z = -6$. Adding this resulting statement to the initial first equation algebraically eliminates the variable $z$, generating $3x + 6y = -6$. Pairing this derived statement with the third original equation creates a simplified two-variable system: $\left\{\begin{array}{l} 3x + 6y = -6 \\ 2x + 3y = -5 \end{array}\right.$. To solve for $x$, multiply the second equation by $-2$ to obtain $-4x - 6y = 10$ and add it to the first to cleanly eliminate $y$, yielding $-x = 4$, which means $x = -4$. Sequentially substituting $x = -4$ back into the third equation calculates $y = 1$. Finally, inserting the initial found value $x = -4$ into the first equation allows for solving the remaining variable, resulting in $z = -3$. After verifying these substituted parameters independently against all three original statements, the absolute solution reliably forms the ordered triple $(-4, 1, -3)$.