I have some problems trying to find the number of solutions for this linear system (with $\lambda \in \Bbb R$):
$$ \begin{cases} x+y+2z+3t=\lambda \\ y+z+\lambda t = -1 \\ x+2y+3z+4t=0 \end{cases} $$
or:
$$ A = \begin{bmatrix} 1 & 1 & 2 & 3 \\ 0 & 1 & \lambda & \lambda \\ 1 & 2 & 3 & 4 \\ \end{bmatrix} $$ and: $$ b= \begin{bmatrix} \lambda \\ -1 \\ 0 \\ \end{bmatrix} $$
At a first glance I noticed that $rank(A)$ should be at least 2 $\forall \lambda \in \Bbb R$ since:
$$ M_2= \begin{bmatrix} 1 & 1\\ 0 & 1\\ \end{bmatrix}, det(M_2)\neq0 $$
and also that if $\lambda = 1$:
$$ A = \begin{bmatrix} 1 & 1 & 2 & 3 \\ 0 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ \end{bmatrix} , b= \begin{bmatrix} 1 \\ -1 \\ 0 \\ \end{bmatrix} $$ then: $$r_3 = r_1+r_2$$ and then: $$rank(A) = rank(Ab) = 2$$ For Rouche-Capelli theorem the system should have $\infty^2$ solutions: $$ y=t+x-3, z=2-2t-x $$
then if $\lambda = 0:$
$$ A = \begin{bmatrix} 1 & 1 & 2 & 3 \\ 0 & 1 & 0 & 0 \\ 1 & 2 & 3 & 4 \\ \end{bmatrix} , b= \begin{bmatrix} 0 \\ -1 \\ 0 \\ \end{bmatrix} $$
we have: $$ M_3= \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 3 \\ \end{bmatrix}, det(M_3) \neq0 $$
$$ rank(A)=rank(Ab)=3 $$ For Rouche-Capelli theorem the system should have $\infty$ solutions: $$ x=-t-1, y=-1, z=1-t $$
but now I have no more ideas .. and I'm long from having completed my task. Any hints?
Assuming you have a typo (as in the comment above) your system becomes:
$$ \begin{cases} x+y+2z+3t=\lambda \\ y+z+\lambda t = -1 \\ x+2y+3z+4t=0 \end{cases} $$
I am assuming that $\lambda$ is a constant.
Write the last equation as:
$$x+y+2z+3t+y+z+t=0$$
From equation (1) this is $=\lambda$, so it becomes
$$(x+y+2z+3t)+y+z+t=0$$
Which is:
$$y+z+t=-\lambda$$
Use this value in equation (1) to get:
$$x+z+2t-\lambda=\lambda$$
So the system becomes:
$$ \begin{cases} x+z+2t=2\lambda \\ y+z+\lambda t = -1 \\ y+z+t=-\lambda \end{cases} $$
Manipulating equations (2) and (3) in the above system results (assuming $\lambda$ is not equal 1): $t=1$
When you plug this value in you get the system: $$ \begin{cases} x+z=2\lambda-2 \\ y+z=-\lambda -1 \\ y+z=-\lambda -1 \end{cases} $$
The last equation is redundant. So we have 2 equations and 3 variables. You may choose to use $x=ky$ where $k$ is a nonzero constant. to get a system of $2$ equations and $2$ variables. This will lead to a infinite number of solutions since k can assume any non-zero value.