Given the vectors: \[ \mathbf{a} = 2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}, \quad \mathbf{b} = -\mathbf{i} + 2\mathbf{j} + \mathbf{k}, \quad \mathbf{c} = 3\mathbf{i} + \mathbf{j} \] We need to find the value of \( t \) such that \( \mathbf{a} + t\mathbf{b} \) is perpendicular to \( \mathbf{c} \). \[ (\mathbf{a} + t\mathbf{b}) \cdot \mathbf{c} = 0 \] Expanding the dot product: \[ \mathbf{a} \cdot \mathbf{c} + t \mathbf{b} \cdot \mathbf{c} = 0 \] Solving for \( t \): \[ t = -\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{b} \cdot \mathbf{c}} \] Calculating the dot products: \[ \mathbf{a} \cdot \mathbf{c} = (2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) \cdot (3\mathbf{i} + \mathbf{j}) = 2 \cdot 3 + 2 \cdot 1 + 3 \cdot 0 = 6 + 2 + 0 = 8 \] \[ \mathbf{b} \cdot \mathbf{c} = (-\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \cdot (3\mathbf{i} + \mathbf{j}) = -1 \cdot 3 + 2 \cdot 1 + 1 \cdot 0 = -3 + 2 + 0 = -1 \] Substituting the values: \[ t = -\frac{8}{-1} = 8 \] Therefore, the value of \( t \) is: \[ \boxed{8} \]