Question:

Step 1: Set up the collinearity condition. For points $A_1, B_1, C_1$ to be collinear, vectors $\vec{A_1B_1}$ and $\vec{A_1C_1}$ must be parallel. $\vec{A_1B_1} = \vec{b_1} - \vec{a_1} = (2\vec{a} + \lambda\vec{b}) - (\lambda\vec{a} + 3\vec{b}) = (2-\lambda)\vec{a} + (\lambda-3)\vec{b}$ $\vec{A_1C_1} = \vec{c_1} - \vec{a_1} = (\vec{a} + \vec{b}) - (\lambda\vec{a} + 3\vec{b}) = (1-\lambda)\vec{a} + (1-3)\vec{b} = (1-\lambda)\vec{a} - 2\vec{b}$ Step 2: Apply the collinearity condition. For collinearity, $\vec{A_1B_1} = k\vec{A_1C_1}$ for some scalar $k \neq 0$: $(2-\lambda)\vec{a} + (\lambda-3)\vec{b} = k[(1-\lambda)\vec{a} - 2\vec{b}]$ $(2-\lambda)\vec{a} + (\lambda-3)\vec{b} = k(1-\lambda)\vec{a} - 2k\vec{b}$ Step 3: Equate coefficients. Since $\vec{a}$ and $\vec{b}$ are non-collinear, coefficients must match independently: For $\vec{a}$: $2-\lambda = k(1-\lambda)$ ... (1) For $\vec{b}$: $\lambda - 3 = -2k$ ... (2) Step 4: Solve for $\lambda$. From equation (2): $k = \frac{3-\lambda}{2}$ Substitute into equation (1): $2-\lambda = \frac{3-\lambda}{2}(1-\lambda)$ $2(2-\lambda) = (3-\lambda)(1-\lambda)$ $4 - 2\lambda = 3 - 3\lambda - \lambda + \lambda^2$ $4 - 2\lambda = 3 - 4\lambda + \lambda^2$ $\lambda^2 - 2\lambda - 1 = 0$ Step 5: Find the sum of roots. Using Vieta's formula for $\lambda^2 - 2\lambda - 1 = 0$: $\text{Sum of roots} = -\frac{(-2)}{1} = 2$ ∴ Answer: 2