There are some problems here. You have computed, correctly,
$$\begin{vmatrix}
\lambda & 1 & 1\\
1& \lambda & 1\\
1 & 1 & \lambda
\end{vmatrix}=\lambda^3-3\lambda + 2.$$
However, this is not $0$ if and only if $\lambda = 0$. Indeed, substituting $\lambda = 0$ into this produces a determinant of $2$.
Further, if the determinant is non-zero, as you say, you'll get a unique solution. This means you can't use $z$ (or any of the variables) as a free parameter. The solution is unique, hence there must be one and only one value of $z$ that produces a solution.
Let's look at the polynomial $\lambda^3-3\lambda + 2$. We can solve this via the usual methods: guess a root, use the factor theorem, then solve the remaining quadratic. If we look at the matrix itself, it jumps out at me that $\lambda = 1$ must be a root, as the corresponding matrix would have all the same rows/columns, making the rows/columns linearly dependent. So, I expect $\lambda = 1$ to be a root (it is, check it!), and $\lambda - 1$ to be a factor. Using polynomial division,
$$\lambda^3 - 3\lambda + 2 = (\lambda - 1)(\lambda^2 + \lambda - 2).$$
Solving/factorising the quadratic by whatever means you prefer, you should then get
$$\lambda^3 - 3\lambda + 2 = (\lambda - 1)^2(\lambda - 2).$$
While this information will be important, it will not, by itself, help us solve the system in place. Cramer's Rule could be used when $\lambda \neq -2, 1$, but just because the system doesn't have a unique solution doesn't mean it can't be solved.
Instead, just set up an augmented matrix, as usual:
$$\left[\begin{array}{ccc|c}
\lambda & 1 & 1 & 1 \\
1 & \lambda & 1 & \mu \\
1 & 1 & \lambda & \mu^2
\end{array}\right].$$
We can now perform all the usual elementary row operations. Remember, $\lambda$ is just a scalar. Though, we must be careful when dividing a row by a scalar expression in terms of $\lambda$, in case that expression happens to be $0$. First, let's swap row $1$ and $3$, to get $1$ in the top left:
$$\left[\begin{array}{ccc|c}
1 & 1 & \lambda & \mu^2 \\
1 & \lambda & 1 & \mu \\
\lambda & 1 & 1 & 1
\end{array}\right].$$
Subtract row $1$ from $2$, and $\lambda$ lots of row $1$ from row $3$:
$$\left[\begin{array}{ccc|c}
1 & 1 & \lambda & \mu^2 \\
0 & \lambda - 1 & 1 - \lambda & \mu - \mu^2 \\
0 & 1 - \lambda & 1 - \lambda^2 & 1 - \lambda\mu^2
\end{array}\right]. \tag{1}$$
The next step I'd want to do is divide a row by $\lambda - 1$ or $1 - \lambda$. But, what if this scalar is $0$, i.e. if $\lambda = 1$? Then the matrix turns into:
$$\left[\begin{array}{ccc|c}
1 & 1 & 1 & \mu^2 \\
0 & 0 & 0 & \mu - \mu^2 \\
0 & 0 & 0 & 1 - \mu^2
\end{array}\right],$$
which can only have a solution if $1 - \mu^2 = 0$ and $\mu - \mu^2 = 0$. The first equation is satisfied if and only if $\mu = \pm 1$, while the second is satisfied if and only if $\mu = 0$ or $1$. Thus, $\mu = 1$ is the only possible value that produces solution. In this case, we get one non-zero row:
$$x + y + z = 1,$$
which we solve by making $y$ and $z$ both free parameters.
Moving back to $(1)$, we can now divide rows $2$ and $3$ by $1 - \lambda$, giving us,
$$\left[\begin{array}{ccc|c}
1 & 1 & \lambda & \mu^2 \\
0 & -1 & 1 & \frac{\mu - \mu^2}{1 - \lambda} \\
0 & 1 & \lambda + 1 & \frac{1 - \lambda\mu^2}{1 - \lambda}
\end{array}\right].$$
Add row $2$ to row $3$, then multiply row $2$ by $-1$ to get
$$\left[\begin{array}{ccc|c}
1 & 1 & \lambda & \mu^2 \\
0 & 1 & -1 & \frac{\mu^2 - \mu}{1 - \lambda} \\
0 & 0 & \lambda + 2 & \frac{1 - \lambda\mu^2 + \mu - \mu^2}{1 - \lambda}
\end{array}\right].$$
Again, we find ourselves wanting to divide by the scalar $\lambda + 2$, but we don't know if it's $0$. Try substituting $\lambda = -2$ into the matrix, and find the two values of $\mu$ that produce solutions. This time, when $\mu$ is one of those two values, there will only be a one-parameter solution.
Otherwise, $\lambda + 2 \neq 0$, and we get
$$\left[\begin{array}{ccc|c}
1 & 1 & \lambda & \mu^2 \\
0 & 1 & -1 & \frac{\mu^2 - \mu}{1 - \lambda} \\
0 & 0 & 1 & \frac{1 - \lambda\mu^2 + \mu - \mu^2}{(1 - \lambda)(\lambda + 2)}
\end{array}\right].$$
This gives us immediately our unique value of $z$ (assuming $\lambda \neq 1, -2$), and the other values can be found through back-substitution.
It's a long question for sure, but it can be done with all the elementary techniques.
