equation system: which $k$ will yield infinitely many solutions?

Question

Determine such values ​​of k that the equation system has infinitely many solutions. How many parameters these solutions depend on. $\begin{cases}x+2y-3z+t=1\\x+4y+3z+4t=-4\\x-4y-21z-8t=k\end{cases}$

$\begin{bmatrix}1&2&-3&1|&1\\1&4&3&4|&-4\\1&-4&-21&-8|&k\end{bmatrix}$...after transformations i got $\begin{bmatrix}1&2&-3&1|&1\\0&1&3&\frac{3}{2}|&-\frac{5}{2}\\0&0&1&0|&\frac{k-16}{-6}\end{bmatrix}$
I don't know what to do next to finish the task

Answer 1

If $a(x+2y-3z+t)+b(x+4y+3z+4t)=x-4y-21z-8t,$

then $a+b=1$ and $2a+4b=-4$, so $a=\color{blue}4$ and $b=\color{purple}{-3}$.

If you multiply the first equation by $\color{blue}4$ and the second equation by $\color{purple}{-3}$ and add them,

you will get the third equation when $k=\,?$

Answer 2

Hint:

1. When will the system have infinite solutions rather than no solution? Or, when will $k$ cause inconsistency of the system ($0=\text{non-zero number}$, which is impossible)? (Think: Why there cannot be a unique solution?)
2. After finding $k$ that assures the system has infinite solutions, you can solve equations by setting free parameters as usual.