$\begin{bmatrix} 1 &1 & 2 & 2\\ 1 & 1 &1 & 3\\ a& b& b& 1 \end{bmatrix}$
First observe the matrix, this matrix will have rank 2 meaning only two independent row and columns are there. We use row approach as it is easier for this matrix.
Linear dependent rows means we can write it as combination of other rows, therefore we can write $R_{3}$ as some linear combination of $R_{1}$ and $R_{2}$. this also consequently means that i can make $R_{3}$ zero using row operation
Another hint in question is presence of 1 in $R_{3}$, to get $R_{3}$ as zero we need to make 1 zero, so what can be used to make it zero?
Using $R_{1}$ and $R_{2}$, say we subtract x$R_{1}$+y$R_{2}$from $R_{3}$, we end up with
$\begin{bmatrix} 1 &1 & 2 & 2\\ 1 & 1 &1 & 3\\ a-(x+y)& b-(x+y)& b-(2x+y)& 1-(2x+3y) \end{bmatrix}$.
Its obvious that b-(x+y) and b-(2x+y) can only be zero together when x is zero, also 1-(2x+3y) is zero when y = 1/3.
subtract $R_{2}$/3 from $R_{3}$, we end up with $\begin{bmatrix} 1 &1 & 2 & 2\\ 1 & 1 &1 & 3\\ a-1/3& b-1/3& b-1/3& 0 \end{bmatrix}$
this gives us a=1/3 and b = 1/3.