1. **Problem Statement:** Find the sum of all values of $\beta$ for which the points represented by position vectors $$\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}, \quad \vec{B} = 2\hat{i} + (3-\beta)\hat{j} + 2\hat{k}, \quad \vec{C} = 4\hat{i} + 4\hat{j} + 2\hat{k}, \quad \vec{D} = (8-\beta)\hat{i} + 5\hat{j} + 4\hat{k}$$ are coplanar. 2. **Formula and Concept:** Four points are coplanar if the scalar triple product of vectors formed by three of them is zero. We can use vectors: $$\vec{AB} = \vec{B} - \vec{A}, \quad \vec{AC} = \vec{C} - \vec{A}, \quad \vec{AD} = \vec{D} - \vec{A}$$ The points are coplanar if: $$\vec{AB} \cdot (\vec{AC} \times \vec{AD}) = 0$$ 3. **Calculate vectors:** $$\vec{AB} = (2-2)\hat{i} + ((3-\beta)-3)\hat{j} + (2-1)\hat{k} = 0\hat{i} + (-\beta)\hat{j} + 1\hat{k}$$ $$\vec{AC} = (4-2)\hat{i} + (4-3)\hat{j} + (2-1)\hat{k} = 2\hat{i} + 1\hat{j} + 1\hat{k}$$ $$\vec{AD} = ((8-\beta)-2)\hat{i} + (5-3)\hat{j} + (4-1)\hat{k} = (6-\beta)\hat{i} + 2\hat{j} + 3\hat{k}$$ 4. **Compute cross product $\vec{AC} \times \vec{AD}$:** $$\vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 6-\beta & 2 & 3 \end{vmatrix} = \hat{i}(1 \cdot 3 - 1 \cdot 2) - \hat{j}(2 \cdot 3 - 1 \cdot (6-\beta)) + \hat{k}(2 \cdot 2 - 1 \cdot (6-\beta))$$ $$= \hat{i}(3 - 2) - \hat{j}(6 - (6 - \beta)) + \hat{k}(4 - (6 - \beta)) = \hat{i}(1) - \hat{j}(6 - 6 + \beta) + \hat{k}(4 - 6 + \beta)$$ $$= \hat{i} - \hat{j}(\beta) + \hat{k}(\beta - 2)$$ 5. **Compute scalar triple product:** $$\vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (0)(1) + (-\beta)(-\beta) + (1)(\beta - 2) = 0 + \beta^2 + \beta - 2$$ 6. **Set scalar triple product to zero for coplanarity:** $$\beta^2 + \beta - 2 = 0$$ 7. **Solve quadratic equation:** $$\beta^2 + \beta - 2 = 0$$ Using factorization: $$ (\beta + 2)(\beta - 1) = 0 $$ So, $$ \beta = -2 \quad \text{or} \quad \beta = 1 $$ 8. **Sum of all values of $\beta$:** $$ n = (-2) + 1 = -1 $$ 9. **Absolute value:** $$ |n| = |-1| = 1 $$ **Final answer:** 1