Let $M$ be a point in the interior of triangle $ABC$. Let $A'$ lie on $BC$ with $MA'$ perpendicular to $BC$. Define $B'$ on $CA$ and $C'$ on $AB$ similarly.  Define

\[
p(M) = \frac{MA' \cdot MB' \cdot MC'}{MA \cdot MB \cdot MC}.
\]

Determine, with proof, the location of $M$ such that $p(M)$ is maximal.  Let $\mu(ABC)$ denote this maximum value.  For which triangles $ABC$ is the value of $\mu(ABC)$ maximal?
1. **Define the angles and relationships:**
   Let \( \alpha = \angle AMB' \), \( \beta = \angle BMC' \), and \( \gamma = \angle CMA' \). We have the following relationships:
   \[
   MA = MC' \csc \alpha, \quad MB = MC' \sin \beta, \quad MC = MA' \csc \gamma
   \]
   Therefore, the expression for \( p(M) \) becomes:
   \[
   p(M) = \frac{MA' \cdot MB' \cdot MC'}{MA \cdot MB \cdot MC} = \sin \alpha \sin \beta \sin \gamma
   \]

2. **Maximize the product of sines:**
   We need to maximize \( \sin \alpha \sin \beta \sin \gamma \). By the Trigonometric form of Ceva's Theorem, we have:
   \[
   \sin \alpha \sin \beta \sin \gamma = \sin (A - \alpha) \sin (B - \beta) \sin (C - \gamma)
   \]
   where \( A = \angle CAB \), \( B = \angle ABC \), and \( C = \angle BCA \).

3. **Simplify the expression:**
   We seek to maximize:
   \[
   \sin \alpha \sin(A - \alpha) \sin \beta \sin(B - \beta) \sin \gamma \sin(C - \gamma)
   \]
   Using the product-to-sum identities, we can rewrite this as:
   \[
   \sin \alpha \sin(A - \alpha) = \frac{1}{2} [\cos(A - 2\alpha) - \cos A]
   \]
   Similarly for the other terms, we get:
   \[
   \sin \alpha \sin(A - \alpha) \sin \beta \sin(B - \beta) \sin \gamma \sin(C - \gamma) = \frac{1}{8} (\cos(A - 2\alpha) - \cos A)(\cos(B - 2\beta) - \cos B)(\cos(C - 2\gamma) - \cos C)
   \]

4. **Analyze the bounds:**
   Since \( 0 \leq \cos(A - 2\alpha) - \cos A \leq 1 - \cos A \), and similarly for the other terms, we have:
   \[
   \frac{1}{8} (\cos(A - 2\alpha) - \cos A)(\cos(B - 2\beta) - \cos B)(\cos(C - 2\gamma) - \cos C) \leq \frac{(1 - \cos A)(1 - \cos B)(1 - \cos C)}{8}
   \]
   Equality holds if and only if \( \alpha = \frac{A}{2} \), \( \beta = \frac{B}{2} \), and \( \gamma = \frac{C}{2} \). This implies that the point \( M \) that maximizes \( p(M) \) is the incenter of the triangle.

5. **Maximize \( p(M) \) for specific triangles:**
   To find the maximum value of \( p(M) \), we consider:
   \[
   8p(M)^2 = (1 - \cos A)(1 - \cos B)(1 - \cos C)
   \]
   Let \( f(x) = \ln(1 - \cos x) \). Since \( f''(x) = \frac{1}{\cos x - 1} < 0 \), \( f \) is concave. By Jensen's inequality:
   \[
   f(A) + f(B) + f(C) \leq 3f \left( \frac{A + B + C}{3} \right)
   \]
   This implies:
   \[
   (1 - \cos A)(1 - \cos B)(1 - \cos C) \leq \left(1 - \cos \frac{\pi}{3}\right)^3
   \]
   Equality holds if and only if \( A = B = C \), i.e., the triangle is equilateral.

Conclusion:
The point \( M \) that maximizes \( p(M) \) is the incenter of the triangle \( ABC \). The value of \( \mu(ABC) \) is maximized when the triangle is equilateral.

The final answer is \( \boxed{ \mu(ABC) } \) is maximized when \( ABC \) is an equilateral triangle.