Question:

Step 1: Set up the condition for non-trivial solutions. For a system of homogeneous linear equations to have a non-trivial solution, the determinant of the coefficient matrix must equal zero. We write: $$\Delta = \begin{vmatrix} \lambda-1 & 3\lambda+1 & 2\lambda \\ \lambda-1 & 4\lambda-2 & \lambda+3 \\ 2 & 3\lambda+1 & 3(\lambda-1) \end{vmatrix} = 0$$ Step 2: Expand and solve the determinant equation. Computing the determinant and setting it equal to zero yields specific values of $\lambda$. Through expansion and simplification, we find that the smallest value of $\lambda$ is $\lambda = 0$. Step 3: Substitute $\lambda = 0$ into the original equations. When $\lambda = 0$, the three equations become: $$(-1)x + (1)y + 0 \cdot z = 0 \quad \Rightarrow \quad -x + y = 0$$ $$(-1)x + (-2)y + (3)z = 0 \quad \Rightarrow \quad -x - 2y + 3z = 0$$ $$2x + (1)y + (-3)z = 0 \quad \Rightarrow \quad 2x + y - 3z = 0$$ Step 4: Solve the system using the first equation. From the first equation: $$-x + y = 0 \implies x = y$$ Step 5: Substitute $x = y$ into the second equation. Substituting $x = y$ into the second equation: $$-x - 2x + 3z = 0$$ $$-3x + 3z = 0$$ $$z = x$$ Step 6: Verify the solution in the third equation. Substituting $x = y$ and $z = x$ into the third equation: $$2x + x - 3x = 0$$ $$0 = 0$$ ✓ This confirms our solution is consistent. Step 7: State the final answer. Since $x = y = z$, we have: $$x : y : z = 1 : 1 : 1$$ The answer is **(Option 4): $1:1:1$**