If the system of equations: $kx+y+2z=1,3x-y-2z=2$ and $-2x-2y-4z=3$ has infinitely many solutions, then $k$ is equal to

Find the value of $k$:

The system of equations will have infinitely many solutions when

$D=D_1=D_2=D_3=0$, that is:

$\left|\begin{array}{ccc}k& 1& 2\\ 3& -1& -2\\ -2& -2& -4\end{array}\right|=\left|\begin{array}{ccc}1& 1& 2\\ 2& -1& -2\\ 3& -2& -4\end{array}\right|=\left|\begin{array}{ccc}k& 1& 2\\ 3& 2& -2\\ -2& 3& -4\end{array}\right|=\left|\begin{array}{ccc}k& 1& 1\\ 3& -1& 2\\ -2& -2& 3\end{array}\right|=0$

Now solving $D_2=0$

$$\begin{gathered} \left|\begin{array}{ccc}k& 1& 2\\ 3& 2& -2\\ -2& 3& -4\end{array}\right|=0 \\ \Rightarrow k\left[\left(-8\right)-\left(-6\right)\right]-1\left[-12-4\right]+2\left[9-\left(-4\right)\right]=0 \\ \Rightarrow k\left(-2\right)+16+26=0 \\ \Rightarrow -2k=-42 \\ \Rightarrow k=21 \end{gathered}$$

So, the value of $k$ is $21$.

Hence, the answer is $21$.