Solution: The given system of equation is \begin{equation}\label{eq:25May2025-1} \begin{gathered} x + 5y - z = 1 \\ 4x + 3y - 3z = 7 \\ 24 x + y + \lambda z = \mu. \end{gathered} \end{equation} The system \eqref{eq:25May2025-1} has infinitely many solutions, so the rank of the coefficient matrix must be less than $3$ and it should be equal to that of augmented matrix. Let us do the row reduction. The augmented matrix is given by \begin{align*} \begin{bmatrix} 1 & 5 & -1 & 1 \\ 4 & 3 & -3 & 7 \\ 24 & 1 & \lambda & \mu \\ \end{bmatrix} & \xrightarrow[R_3 \rightarrow R_3 - 24 R_1]{R_2 \rightarrow R_2 - 4R_1} \begin{bmatrix} 1 & 5 & -1 & 1 \\ 0 & -17 & 1 & 3 \\ 0 & -119 & \lambda +24 & \mu -24 \\ \end{bmatrix} \\[2ex] & \xrightarrow{R_2 \leftrightarrow - R_2} \begin{bmatrix} 1 & 5 & -1 & 1 \\ 0 & 17 & -1 & -3 \\ 0 & -119 & \lambda +24 & \mu -24 \\ \end{bmatrix} \\[2ex] & \xrightarrow{R_3 \rightarrow R_3 - + 7 R_2} \begin{bmatrix} 1 & 5 & -1 & 1 \\ 0 & 17 & -1 & -3 \\ 0 & 0 & \lambda + 17 & \mu - 27 \\ \end{bmatrix}. \end{align*}

In order to have infinitely many solutions, we must have \[ \lambda + 17 = 0 \text{ and } \mu - 27 = 0 \implies \lambda = -17 \text{ and } \mu =27. \] Therefore the solution is \begin{align*} z = 3 + 17y \text{ and } x = 1 + z - 5y = 4 + 12y. \end{align*} Hence, the set of solution is \begin{align*} \mathcal{S} = \left\{\begin{bmatrix} 4 + 12y \\ y \\ 3 + 17y \\ \end{bmatrix} : y \in \mathbb{R} \right\} . \end{align*}

Now we will work on the restrictive condition, that is integral solutions with \begin{align*} 7 \leq x + y + z \leq 77 & \implies 7 \leq 7 + 30y \leq 77 \\ & \implies 0 \leq 30 y \leq 70 \\ & \implies 0 \leq y \leq \frac{7}{3} = 2\frac{1}{3}. \end{align*} Therefore, the number of integral solutions with $7 \leq x + y + z \leq 77$ is $3$.