In a triangle, the ratio of the interior angles is $1 : 5 : 6$, and the longest
side has length $12$. What is the length of the altitude (height) of the triangle that
is perpendicular to the longest side?
1. **Determine the angles of the triangle:**
   Given the ratio of the interior angles is \(1 : 5 : 6\), let the angles be \(x\), \(5x\), and \(6x\). Since the sum of the interior angles in a triangle is \(180^\circ\), we have:
   \[
   x + 5x + 6x = 180^\circ
   \]
   \[
   12x = 180^\circ
   \]
   \[
   x = 15^\circ
   \]
   Therefore, the angles of the triangle are \(15^\circ\), \(75^\circ\), and \(90^\circ\).

2. **Identify the type of triangle:**
   Since one of the angles is \(90^\circ\), the triangle is a right triangle. The angles are \(15^\circ\), \(75^\circ\), and \(90^\circ\).

3. **Determine the sides of the triangle:**
   In a \(15^\circ-75^\circ-90^\circ\) triangle, the sides are in the ratio \(1 : 2 : \sqrt{3}\). Given that the longest side (the hypotenuse) is \(12\), we can set up the following proportions:
   \[
   \text{Hypotenuse} = 12
   \]
   \[
   \text{Opposite to } 15^\circ = 12 \sin 15^\circ
   \]
   \[
   \text{Adjacent to } 15^\circ = 12 \cos 15^\circ
   \]

4. **Calculate the altitude (height) of the triangle:**
   The altitude is perpendicular to the longest side (the hypotenuse). In a right triangle, the altitude to the hypotenuse can be found using the formula:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Here, the base is the hypotenuse, and the height is the altitude we need to find. The area can also be calculated using the other two sides:
   \[
   \text{Area} = \frac{1}{2} \times 12 \sin 15^\circ \times 12 \cos 15^\circ
   \]
   Equating the two expressions for the area:
   \[
   \frac{1}{2} \times 12 \times h = \frac{1}{2} \times 12 \sin 15^\circ \times 12 \cos 15^\circ
   \]
   Simplifying:
   \[
   12h = 12 \sin 15^\circ \cos 15^\circ
   \]
   \[
   h = \sin 15^\circ \cos 15^\circ
   \]
   Using the double-angle identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have:
   \[
   \sin 30^\circ = 2 \sin 15^\circ \cos 15^\circ
   \]
   Since \(\sin 30^\circ = \frac{1}{2}\), we get:
   \[
   \frac{1}{2} = 2 \sin 15^\circ \cos 15^\circ
   \]
   \[
   \sin 15^\circ \cos 15^\circ = \frac{1}{4}
   \]
   Therefore:
   \[
   h = 12 \times \frac{1}{4} = 3
   \]

The final answer is \(\boxed{3}\)