I'd like to discuss number of solutions based on value of parameter $a$.
$\begin{cases} (a+3)x-4y+2z+2t=a+1\\ 2x-2y-z+t=a \\ x-2y+z+(a+1)t=a \end{cases}$
Some guidance on how to count rank here would be benficial. I tried to use Gauss elimination method, but I didn't manage to make a staircase-like pattern.
The criterion for solutions to exist is that the matrix of the system and the augmented matrix have the same rank. When this condition is satisfied, the rank is the codimension of the affine space of solutions.
Here, one can conclude without performing a full RREF: \begin{align} &\left[\!\! \begin{array}{cccc|c} a+3&-4&2&2&a+1 \\ 2&-2&-1&1& a \\ 1&-2&1&a+1&a \end{array} \!\!\right]\rightsquigarrow \left[\!\! \begin{array}{cccc|c} 1&-2&1&a+1&a\\ 2&-2&-1&1& a \\ a+3&-4&2&2&a+1 \end{array} \!\!\right]\rightsquigarrow \\[1ex] &\left[\!\! \begin{array}{cccc|c} 1&-2&1&a+1&a\\ 0&2&-3& -2a-1 &-a \\ 0&2(a+1)&-a-1&-a^2-4a-1&-a^2-2a+1 \end{array} \!\!\right]\rightsquigarrow \\[1ex] &\left[\!\! \begin{array}{cccc|c} 1&-2 & 1 & a+1 & a\\ 0 & 2 & -3 & 1 & -2a-1 \\ 0&0&2(a+1) & a^2-a & 1-a \end{array} \!\!\right] \end{align} and clearly, both matrices have rank $3$.