Three planes have the following equations...

$\displaystyle x+2y+z = 3$

$\displaystyle 3x+4y+5z=k$

$\displaystyle 2x-y+kz=1$

$\displaystyle \[

\emph{Augumented Matrix} =

\left( {\begin{array}{cccc}

1 & 2 & 1 & 3 \\

3 & 4 & 5 & k \\

2 & -1 & k & 1

\end{array} } \right)

\]$

$\displaystyle \[

\emph{Reduced - Row echelon form} =

\left( {\begin{array}{cccc}

1 & 0 & 0 & k+\frac{3}{2} \\

0 & 1 & 0 & \frac{-k}{2}+2 \\

0 & 0 & 1 & \frac{-5}{2}

\end{array} } \right)

\]$

$\displaystyle \emph{Find all values of k for which the system of linear equations}$

$\displaystyle \emph{(i) has infinitely many solutions}$

$\displaystyle \emph{(ii) has a unique solution}$

$\displaystyle \emph{(iii) has no solutions}$

Not sure on how to approach this question..