1. For a system of $3$ linear equations in $3$ variables to have infinitely many solutions, we need

   • the coefficient matrix to be singular: $\det(A)=0$,
   • and the equations to be consistent, i.e. the augmented matrix must have the same rank as the coefficient matrix, both $<3$.

2. Write the coefficient matrix: $A=\begin{pmatrix}2&7&\lambda\\[4pt]3&2&5\\[4pt]1&\mu&32\end{pmatrix}$

   Since infinitely many solutions are given, first set $\det(A)=0.$

3. Compute the determinant: $\det(A)=2\begin{vmatrix}2&5\\ \mu&32\end{vmatrix}-7\begin{vmatrix}3&5\\ 1&32\end{vmatrix}+\lambda\begin{vmatrix}3&2\\ 1&\mu\end{vmatrix}$

   $=2(64-5\mu)-7(96-5)+\lambda(3\mu-2)$

   $=128-10\mu-637+\lambda(3\mu-2)$

   $=\lambda(3\mu-2)-10\mu-509$

   Hence, $\lambda(3\mu-2)-10\mu-509=0 \qquad (1)$

4. For infinitely many solutions, the third equation must be a linear combination of the first two (since rank must be $2$).

   So let $R_3=aR_1+bR_2$ where \begin{align*} a(2,7,\lambda\mid 3)+b(3,2,5\mid 4)=(1,\mu,32\mid -1). \end{align*}

   Equating coefficients gives: \begin{align*} 2a+3b&=1 \qquad (2)\ 3a+4b&=-1 \qquad (from constants) \end{align*}

   Solve these two equations:

   From $2a+3b=1$ and $3a+4b=-1,$

   multiply first by $3$ and second by $2$: \begin{align*} 6a+9b&=3\ 6a+8b&=-2 \end{align*} Subtracting, $b=5.$ Then from $2a+3b=1$, $2a+15=1 \Rightarrow 2a=-14 \Rightarrow a=-7.$

5. Now use the $y$-coefficient and $z$-coefficient:

   $\mu=7a+2b=7(-7)+2(5)=-49+10=-39$

   $32=\lambda a+5b=-7\lambda+25$ $-7\lambda=7 \Rightarrow \lambda=-1.$

6. Therefore, $\lambda-\mu=(-1)-(-39)=38.$

7. Verification in determinant condition: $\lambda(3\mu-2)-10\mu-509=(-1)(3(-39)-2)-10(-39)-509$ $=(-1)(-119)+390-509=119+390-509=0,$ so the condition is satisfied.

Hence the required integer is $\boxed{38}.$