If $ a$, $ b$, and $ c$ are positive real numbers such that $ a(b \plus{} c) \equal{} 152$, $ b(c \plus{} a) \equal{} 162$, and $ c(a \plus{} b) \equal{} 170$, then abc is

$ \textbf{(A)}\ 672 \qquad
\textbf{(B)}\ 688 \qquad
\textbf{(C)}\ 704 \qquad
\textbf{(D)}\ 720 \qquad
\textbf{(E)}\ 750$
1. Given the equations:
   \[
   a(b + c) = 152, \quad b(c + a) = 162, \quad c(a + b) = 170
   \]
   We start by expanding and adding these equations.

2. Expand each equation:
   \[
   ab + ac = 152
   \]
   \[
   bc + ba = 162
   \]
   \[
   ca + cb = 170
   \]

3. Add the three equations:
   \[
   (ab + ac) + (bc + ba) + (ca + cb) = 152 + 162 + 170
   \]
   Simplify the left-hand side:
   \[
   2(ab + bc + ca) = 484
   \]
   Divide both sides by 2:
   \[
   ab + bc + ca = 242
   \]

4. Now, we need to find \(ab\), \(bc\), and \(ca\). We use the fact that:
   \[
   ab + bc + ca = 242
   \]
   Subtract each of the original equations from this sum:
   \[
   ab + bc + ca - (ab + ac) = 242 - 152 \implies bc = 90
   \]
   \[
   ab + bc + ca - (bc + ba) = 242 - 162 \implies ac = 80
   \]
   \[
   ab + bc + ca - (ca + cb) = 242 - 170 \implies ab = 72
   \]

5. To find \(abc\), we multiply \(ab\), \(bc\), and \(ca\):
   \[
   (ab)(bc)(ca) = (72)(90)(80)
   \]
   Simplify the product:
   \[
   abc \cdot abc = 72 \cdot 90 \cdot 80
   \]
   \[
   (abc)^2 = 72 \cdot 90 \cdot 80
   \]
   Calculate the product:
   \[
   72 \cdot 90 = 6480
   \]
   \[
   6480 \cdot 80 = 518400
   \]
   Therefore:
   \[
   (abc)^2 = 518400
   \]
   Take the square root of both sides:
   \[
   abc = \sqrt{518400} = 720
   \]

The final answer is \(\boxed{720}\)