The vector equation of the line is given by: \[\mathbf{r} = (6, 7, 7) + t(3, 2, -2).\] Let \(\mathbf{P} = (1, 2, 3)\) and the position vector of a point on the line be \(\mathbf{r_0}\). The direction vector of the line is \(\mathbf{d} = (3, 2, -2).The perpendicular distance \(D\) from a point \((x_1, y_1, z_1)\) to a line is calculated using the formula:\[D = \frac{\|\mathbf{d} \times (\mathbf{r_0} - \mathbf{P})\|}{\|\mathbf{d}\|}.\]Step 1: Calculate \((\mathbf{r_0} - \mathbf{P})\):\[\mathbf{r_0} - \mathbf{P} = (6 - 1, 7 - 2, 7 - 3) = (5, 5, 4).\]Step 2: Compute the cross product \(\mathbf{d} \times (\mathbf{r_0} - \mathbf{P})\):\[\mathbf{d} \times (\mathbf{r_0} - \mathbf{P}) = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}
3 & 2 & -2
5 & 5 & 4\end{vmatrix}.\]\[\mathbf{d} \times (\mathbf{r_0} - \mathbf{P}) = \mathbf{i}(2 \cdot 4 - (-2) \cdot 5) - \mathbf{j}(3 \cdot 4 - (-2) \cdot 5) + \mathbf{k}(3 \cdot 5 - 2 \cdot 5).\]\[\mathbf{d} \times (\mathbf{r_0} - \mathbf{P}) = \mathbf{i}(8 + 10) - \mathbf{j}(12 + 10) + \mathbf{k}(15 - 10).\]\[\mathbf{d} \times (\mathbf{r_0} - \mathbf{P}) = (18, -22, 5).\]Step 3: Determine the magnitude of the cross product:\[\|\mathbf{d} \times (\mathbf{r_0} - \mathbf{P})\| = \sqrt{18^2 + (-22)^2 + 5^2} = \sqrt{324 + 484 + 25} = \sqrt{833}.\]Step 4: Determine the magnitude of the direction vector \(\mathbf{d}\):\[\|\mathbf{d}\| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}.\]Step 5: Calculate the distance \(D\):\[D = \frac{\|\mathbf{d} \times (\mathbf{r_0} - \mathbf{P})\|}{\|\mathbf{d}\|} = \frac{\sqrt{833}}{\sqrt{17}} = \sqrt{\frac{833}{17}} = 7.\]