The tetrahedron $ S.ABC$ has the faces $ SBC$ and $ ABC$ perpendicular. The three angles at $ S$ are all $ 60^{\circ}$ and $ SB \equal{} SC \equal{} 1$. Find the volume of the tetrahedron.
1. **Identify the given information and assumptions:**
   - The tetrahedron \( S.ABC \) has faces \( SBC \) and \( ABC \) perpendicular.
   - The three angles at \( S \) are all \( 60^\circ \).
   - \( SB = SC = 1 \).

2. **Determine the geometry of triangle \( SBC \):**
   - Since \( \angle BSC = 60^\circ \) and \( SB = SC = 1 \), triangle \( SBC \) is an equilateral triangle.
   - The area of an equilateral triangle with side length \( a \) is given by:
     \[
     \text{Area} = \frac{\sqrt{3}}{4} a^2
     \]
   - For \( a = 1 \):
     \[
     \text{Area of } \triangle SBC = \frac{\sqrt{3}}{4} \cdot 1^2 = \frac{\sqrt{3}}{4}
     \]

3. **Determine the height of the tetrahedron:**
   - Since faces \( SBC \) and \( ABC \) are perpendicular, \( S \) lies directly above the centroid of \( \triangle ABC \).
   - The centroid of an equilateral triangle divides the height in a ratio of 2:1.
   - The height \( h \) of an equilateral triangle with side length \( a \) is:
     \[
     h = \frac{\sqrt{3}}{2} a
     \]
   - For \( a = 1 \):
     \[
     h = \frac{\sqrt{3}}{2}
     \]
   - The height of the tetrahedron from \( S \) to the plane \( ABC \) is:
     \[
     \text{Height} = \frac{\sqrt{3}}{2}
     \]

4. **Calculate the volume of the tetrahedron:**
   - The volume \( V \) of a tetrahedron is given by:
     \[
     V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
     \]
   - Using the base area of \( \triangle SBC \) and the height from \( S \) to \( ABC \):
     \[
     V = \frac{1}{3} \times \frac{\sqrt{3}}{4} \times \frac{\sqrt{3}}{2}
     \]
   - Simplify the expression:
     \[
     V = \frac{1}{3} \times \frac{3}{8} = \frac{1}{8}
     \]

The final answer is \(\boxed{\frac{1}{8}}\).