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
First, just find the vectors $\mathbf{T}(2)$ and $\mathbf{N}(2)$.
They want you to find the scalars $c_1, c_2$ so that $\mathbf{r}(2) = c_1\mathbf{T}(2) + c_2\mathbf{N}(2)$.
As a hint, what happens if you take the dot product of both sides with $\mathbf{T}(2)$, given that $\mathbf{T}(2)$ and $\mathbf{N}(2)$ are perpendicular? Ditto with $\mathbf{N}(2)$?