Solution
= Solution
Subtracting the second equation from the first gives $2x-\lambda z=-4$. With $\lambda=2$ this is $x-z=-2$, whereas the third equation requires $x-z=-2\mu$. Consistency therefore requires
$$
\boxed{\mu=1.}
$$
At that value one parameter remains free:
$$
\boxed{(x,y,z)=(t-2,\,7-5t,\,t),\qquad t\in\mathbb R.}
$$
Thus the solution is not unique. For any other $\mu$ the two required values of $x-z$ conflict and the <linear system> has no solution.