Let the system of equations $$x + 5y - z = 1$$, $$4x + 3y - 3z = 7$$, $$24x + y + \lambda z = \mu$$, $$\lambda, \mu \in \mathbf{R}$$, have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy $$7 \le x + y + z \le 77$$, is

The given system is

$$\begin{aligned} x+5y-z &= 1 \quad\; -(1)\\ 4x+3y-3z &= 7 \quad -(2)\\ 24x+y+\lambda z &= \mu \quad -(3) \end{aligned}$$

For infinitely many solutions the three equations must be dependent, i.e. equation $$(3)$$ must be a linear combination of $$(1)$$ and $$(2)$$ with the same combination on the right-hand sides.

Let $$\alpha,\,\beta$$ be real numbers such that

$$\alpha(x+5y-z)+\beta(4x+3y-3z)=24x+y+\lambda z \quad -(4)$$

Matching the coefficients of $$x,\;y,\;z$$ in $$(4)$$ gives

$$\begin{aligned} \alpha+4\beta &= 24 \quad -(5)\\ 5\alpha+3\beta &= 1 \quad -(6)\\ -\alpha-3\beta &= \lambda \quad -(7) \end{aligned}$$

Solving $$(5)$$ and $$(6)$$:
Multiply $$(5)$$ by $$5$$ ⇒ $$5\alpha+20\beta=120$$.
Subtract $$(6)$$ ⇒ $$17\beta = 119 \;\Rightarrow\; \beta = 7$$.
From $$(5)$$ ⇒ $$\alpha = 24-4\beta = 24-28 = -4$$.

Then $$(7)$$ gives $$\lambda = (-\alpha-3\beta)=4-21=-17$$.

The right-hand side of $$(3)$$ must also match:
$$\mu = \alpha\cdot1 + \beta\cdot7 = (-4)\cdot1 + 7\cdot7 = -4+49 = 45$$.

Hence infinitely many solutions occur only for
$$\boxed{\lambda=-17,\;\mu = 45}$$.

With these values, equation $$(3)$$ is redundant and the system reduces to $$(1)$$(2). Solve it parametrically.

From $$(1)$$: $$x = 1 - 5y + z \quad -(8)$$.

Substitute $$(8)$$ in $$(2)$$:

$$4(1-5y+z) + 3y - 3z = 7$$ $$4 - 20y + 4z + 3y - 3z = 7$$ $$4 - 17y + z = 7$$ $$\Rightarrow\; z = 3 + 17y \quad -(9)$$.

Insert $$(9)$$ into $$(8)$$:

$$x = 1 - 5y + (3+17y) = 4 + 12y \quad -(10)$$.

Let $$y = t$$ (any integer). Then

$$x = 4 + 12t,\; y = t,\; z = 3 + 17t \quad -(11)$$

The required condition is $$7 \le x+y+z \le 77$$.

Using $$(11)$$:
$$x+y+z = (4+12t)+t+(3+17t) = 7 + 30t$$.

So

$$7 \le 7 + 30t \le 77 \;\Longrightarrow\; 0 \le 30t \le 70$$
$$\Rightarrow\; 0 \le t \le \frac{70}{30} = \frac{7}{3}$$.

Since $$t$$ is an integer, $$t = 0,1,2$$. Thus there are

$$\boxed{3}$$

integer triples $$(x,y,z)$$ satisfying all the given conditions.

Hence the correct option is Option A (3).