= Solution
\b[Source gap / TODO:] the original printed page 4, confirmed visually, omits the unknown <vector>. It literally equates a $3\times3$ <matrix> with a $3\times1$ <vector>, so it does not define a well-formed <system of linear equations>. The following answer is conditional on the usual intended interpretation $A\mathbf x=(1,k^2,k)^T$; the missing factor is not present in the source.
Under that interpretation, the rows satisfy $r_1-2r_2+r_3=0$. Applying this <linear relation> to the right-hand side gives the necessary consistency condition
$$
1-2k^2+k=0,\qquad (2k+1)(k-1)=0.
$$
It is sufficient too: the upper-left $2\times2$ <minor> has <determinant> $4-9=-5\ne0$, so the first two rows are <linearly independent>. The third row is $2r_2-r_1$, and imposes precisely the condition already found. Therefore
$$
\boxed{k=1\quad\text{or}\quad k=-\tfrac12\qquad\text{if the intended equation is }A\mathbf x=(1,k^2,k)^T.}
$$
For completeness, setting $x_3=t$ gives
$$
x_1=\frac{3k^2-2-4t}{5},\qquad x_2=\frac{3-2k^2+t}{5},\qquad t\in\mathbb R,
$$
and direct substitution satisfies all three <linear equations> exactly for those two values of $k$. An unconditional answer to the malformed printed equality awaits a source correction.
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