1. **State the problem:** We are given the parametric form of a line $r$ and a system of two linear equations $s$. We want to analyze the relationship between the line and the system. 2. **Write the line $r$ in parametric form:** Given: $$r \equiv \frac{x+1}{2} = \frac{y-2}{3} = \frac{z}{\alpha} = t$$ This means: $$x+1 = 2t \Rightarrow x = 2t - 1$$ $$y-2 = 3t \Rightarrow y = 3t + 2$$ $$z = \alpha t$$ 3. **Write the system $s$ explicitly:** $$\begin{cases} -x + y + z = 1 \\ x + 2y - z = 2 \end{cases}$$ 4. **Substitute parametric expressions into the system:** Substitute $x=2t-1$, $y=3t+2$, $z=\alpha t$: First equation: $$-(2t-1) + (3t+2) + \alpha t = 1$$ Simplify: $$-2t + 1 + 3t + 2 + \alpha t = 1$$ $$(-2t + 3t + \alpha t) + (1 + 2) = 1$$ $$ (1 + \alpha) t + 3 = 1$$ Second equation: $$ (2t - 1) + 2(3t + 2) - \alpha t = 2$$ Simplify: $$2t - 1 + 6t + 4 - \alpha t = 2$$ $$(2t + 6t - \alpha t) + (-1 + 4) = 2$$ $$(8 - \alpha) t + 3 = 2$$ 5. **Rewrite the system in terms of $t$:** $$\begin{cases} (1 + \alpha) t + 3 = 1 \\ (8 - \alpha) t + 3 = 2 \end{cases}$$ 6. **Solve each equation for $t$:** First: $$(1 + \alpha) t = 1 - 3 = -2$$ $$t = \frac{-2}{1 + \alpha}$$ Second: $$(8 - \alpha) t = 2 - 3 = -1$$ $$t = \frac{-1}{8 - \alpha}$$ 7. **Set the two expressions for $t$ equal to find $\alpha$:** $$\frac{-2}{1 + \alpha} = \frac{-1}{8 - \alpha}$$ Cross multiply: $$-2 (8 - \alpha) = -1 (1 + \alpha)$$ $$-16 + 2\alpha = -1 - \alpha$$ Bring all terms to one side: $$2\alpha + \alpha = -1 + 16$$ $$3\alpha = 15$$ $$\alpha = 5$$ 8. **Interpretation:** For $\alpha = 5$, the parametric line $r$ satisfies both equations in $s$, meaning the line lies in the plane defined by the system or intersects it consistently. **Final answer:** $$\boxed{\alpha = 5}$$