The system of equations you're trying to express as a matrix operation is:
$\sigma\left(
\begin{array}{c}
1\\
-2\\
-4\\
\end{array}
\right)=\left(
\begin{array}{c}
3\\
0\\
-2\\
\end{array}
\right)+\lambda\left(
\begin{array}{c}
1\\
1\\
1\\
\end{array}
\right)+\mu\left(
\begin{array}{c}
2\\
-1\\
-3\\
\end{array}
\right)$
Hence, $
\begin{pmatrix}
1 & 2 & -1 \\
1 & -1 & 2 \\
1 & -3 & 4
\end{pmatrix} \begin{pmatrix}
\lambda \\
\mu \\
\sigma
\end{pmatrix} =\begin{pmatrix}
-3 \\
0 \\
2
\end{pmatrix}$
Row reducing (what you were referring to as sweeping, I believe) will give you solutions for $\lambda, \mu,$ and $\sigma$. In order to find $\textbf{x}$, we use the fact that
$\sigma\begin{pmatrix}
1 \\
-2 \\
-4
\end{pmatrix}=\textbf{x}$
If I understand correctly $\;i,j,k\;$ is just another name for the coordinate axis, so your lines are
$$\begin{align*}&\ell_1: (1,1,1)+s(1,-1,2)=(s+1,\,-s+1,\,2s+1)\;,\;\;s\in\Bbb R\\{}\\&\ell_2: (4,6,1)+t(2,2,1)=(2t+4,\,2t+6,\,t+1)\;,\;\;t\in\Bbb R\end{align*}$$
Comparing both rightmost expressions, we find the intersection point:
$$\begin{cases}t+1=2s+1\implies t=2s\\2t+6=-s+1\implies s=-2t-5\\2t+4=s+1\implies s=2t+3\end{cases}$$
Lines $\;2-3\;$ give us $\;4t=-8\iff t=-2\;$, and then line $\,1\,$ gives $\;s=-1\;$
so the intersection point is $\;(0,2,-1)\;$, which can be then take as the "anchor" point together with both direction vectors of the lines, and the plane is
$$\pi: (0,2,-1)+s(1,-1,2)+t(2,2,1)$$
For the other form find first the (a) normal to the plane by the vectorial product:
$$(1,-1,2)\times(2,2,1) =\begin{vmatrix}i&j&k\\1&\!\!-1&2\\2&2&1\end{vmatrix}=(-5,3,4)$$
So the plane is $\;-5x+3y+4z+d=0\;$. Inputting $\;(0,2,-1)\;$ gives us $\;d=-2\;$ and thus $\;-5x+3y+4z-2=0\;$ is the plane
Best Answer
$$x= \frac {7+y+3z}{2}$$ and $x= -2y-2z$ [rearranging the two equations of $p_1$ and $p_2$]
$$7+y+3z = -4y-4z$$
$$z = \frac {-5y-7}{7}$$
After plugging in this value of $z$ in the first equation for $x$, I got $x= \frac {14-4y}{7}$
Let $y = t$
$$x= \frac {14-4t}{7}$$
$$z= \frac {-7-5t}{7}$$
This gives the equation of the line to be $(2, 0, -1) + t(- \frac {4}{7}, 1, -\frac {5}{7})$ [by separating the $t$ terms and then taking $t$ common]