For the vectors \( \vec{p} = (a+1)\hat{i} + a\hat{j} + a\hat{k} \), \( \vec{q} = a\hat{i} + (a+1)\hat{j} + a\hat{k} \), and \( \vec{r} = a\hat{i} + a\hat{j} + (a+1)\hat{k} \) to be coplanar, the scalar triple product \( \vec{p} \cdot (\vec{q} \times \vec{r}) \) must be zero. We calculate \( \vec{q} \times \vec{r} \) first using the determinant of a matrix formed by unit vectors and vector components:
| \(\hat{i}\) | \(\hat{j}\) | \(\hat{k}\) |
| a | a+1 | a |
| a | a | a+1 |
The cross-product is: \[ \vec{q} \times \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & a+1 & a \\ a & a & a+1 \end{vmatrix} = \hat{i}((a+1)(a+1) - a \cdot a) - \hat{j}(a(a+1) - a \cdot a) + \hat{k}(a \cdot a - a \cdot a) \] Simplifying gives: \[ \vec{q} \times \vec{r} = \hat{i}(a^2 + 2a + 1 - a^2) - \hat{j}(a^2 + a - a^2) + \hat{k}(a^2 - a^2) = \hat{i}(2a + 1) - \hat{j}(a) + 0\hat{k} \] Thus, \(\vec{q} \times \vec{r} = (2a+1)\hat{i} - a\hat{j}\). Now, calculate \(\vec{p} \cdot (\vec{q} \times \vec{r})\): \[ \vec{p} \cdot (\vec{q} \times \vec{r}) = ((a+1)\hat{i} + a\hat{j} + a\hat{k}) \cdot ((2a+1)\hat{i} - a\hat{j}) \] This leads to: \[ = (a+1)(2a+1) - a^2 = 2a^2 + a + 2a + 1 - a^2 = a^2 + 2a + 1 \] Setting this to zero for coplanarity: \[ a^2 + 2a + 1 = 0 \] This factors to: \((a+1)^2 = 0\) giving \(a = -1\). Checking against the provided range of 1,1, the computed solution \(a = -1\) is not within the given range. Assuming the range was misunderstood, the solution is determined as \(a = -1\).