Step 1: Name the data:
The line passes through $A(6, 7, 7)$ and has direction vector $\vec{d} = 3\hat{i} + 2\hat{j} - 2\hat{k}$.
The given point is $P(1, 2, 3)$.
The vector from $A$ to $P$ is $\vec{AP} = -5\hat{i} - 5\hat{j} - 4\hat{k}$.
Step 2: Use projection on the line:
The foot $Q$ is the point where the projection of $\vec{AP}$ on $\vec{d}$ ends.
First compute the needed dot products.
\[ \vec{AP} \cdot \vec{d} = -15 - 10 + 8 = -17 \]
\[ \vec{d} \cdot \vec{d} = 9 + 4 + 4 = 17 \]
Step 3: Locate the foot:
The scalar multiple of $\vec{d}$ that reaches the foot is $\frac{\vec{AP} \cdot \vec{d}}{\vec{d} \cdot \vec{d}} = -1$.
So $\vec{AQ} = -\vec{d}$, and $Q = A - \vec{d}$.
\[ Q = (6 - 3,\; 7 - 2,\; 7 + 2) = (3, 5, 9) \]
Step 4: Length of the perpendicular by Pythagoras:
In the right triangle $APQ$, we have $PQ^2 = AP^2 - AQ^2$.
$AP^2 = 25 + 25 + 16 = 66$ and $AQ^2 = |\vec{d}|^2 = 17$.
\[ PQ^2 = 66 - 17 = 49 \Rightarrow PQ = 7 \]
Step 5: Equation of the perpendicular PQ:
The direction vector is $\vec{PQ} = Q - P = 2\hat{i} + 3\hat{j} + 6\hat{k}$.
Its length is $\sqrt{4 + 9 + 36} = 7$, which matches Step 4.
Using the point $P(1, 2, 3)$:
\[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6} \]
In vector form this is $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu (2\hat{i} + 3\hat{j} + 6\hat{k})$.
Final Answer:
The perpendicular has length 7 units, its foot is $(3, 5, 9)$, and its equation is given below.
\[ \boxed{7 \text{ units};\; \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}} \]