In the coordinate plane, denote by $ S(a)$ the area of the region bounded by the line passing through the point $ (1,\ 2)$ with the slope $ a$ and the parabola $ y\equal{}x^2$. When $ a$ varies in the range of $ 0\leq a\leq 6$, find the value of $ a$ such that $ S(a)$ is minimized.
1. **Equation of the Line:**
   The line passing through the point \((1, 2)\) with slope \(a\) can be written as:
   \[
   y = ax + (2 - a)
   \]
   This is derived from the point-slope form of the line equation \(y - y_1 = m(x - x_1)\).

2. **Intersection Points:**
   To find the area between the line and the parabola \(y = x^2\), we need to find the points of intersection. Set the equations equal to each other:
   \[
   ax + (2 - a) = x^2
   \]
   Rearrange to form a quadratic equation:
   \[
   x^2 - ax + (a - 2) = 0
   \]
   Solve for \(x\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   x = \frac{a \pm \sqrt{a^2 - 4(a - 2)}}{2} = \frac{a \pm \sqrt{a^2 - 4a + 8}}{2}
   \]
   Let \(x_1 = \frac{a - \sqrt{a^2 - 4a + 8}}{2}\) and \(x_2 = \frac{a + \sqrt{a^2 - 4a + 8}}{2}\).

3. **Area Between the Curves:**
   The area \(S(a)\) between the curves from \(x_1\) to \(x_2\) is given by:
   \[
   S(a) = \int_{x_1}^{x_2} (x^2 - (ax + 2 - a)) \, dx = \int_{x_1}^{x_2} (x^2 - ax - 2 + a) \, dx
   \]
   Simplify the integrand:
   \[
   S(a) = \int_{x_1}^{x_2} (x^2 - ax + a - 2) \, dx
   \]

4. **Evaluating the Integral:**
   Compute the integral:
   \[
   \int (x^2 - ax + a - 2) \, dx = \left[ \frac{x^3}{3} - \frac{ax^2}{2} + (a - 2)x \right]_{x_1}^{x_2}
   \]
   Substitute \(x_1\) and \(x_2\):
   \[
   S(a) = \left( \frac{x_2^3}{3} - \frac{ax_2^2}{2} + (a - 2)x_2 \right) - \left( \frac{x_1^3}{3} - \frac{ax_1^2}{2} + (a - 2)x_1 \right)
   \]

5. **Simplifying the Expression:**
   Using symmetry and properties of definite integrals, we can simplify the expression to:
   \[
   S(a) = \frac{1}{6} (a^2 - 4a + 8)^{3/2}
   \]

6. **Minimizing the Area:**
   To find the value of \(a\) that minimizes \(S(a)\), take the derivative \(S'(a)\) and set it to zero:
   \[
   S'(a) = \frac{1}{4} \sqrt{a^2 - 4a + 8} (2a - 4)
   \]
   Set \(S'(a) = 0\):
   \[
   \frac{1}{4} \sqrt{a^2 - 4a + 8} (2a - 4) = 0
   \]
   Since \(\sqrt{a^2 - 4a + 8} \neq 0\) for real \(a\), we solve:
   \[
   2a - 4 = 0 \implies a = 2
   \]

7. **Verification:**
   Verify that \(a = 2\) is a minimum by checking the second derivative or the nature of the function. Since \(S(a) \geq 0\) and the function is continuous, \(a = 2\) is indeed a minimum.

The final answer is \( \boxed{ a = 2 } \).