Gọi \(d':y = kx + a\) là đường thẳng cần tìm.

Vì \(d'\) đi qua \(A\left( {0;\,\,1} \right)\) nên ta có: \(1 = k.0 + a \Leftrightarrow a = 1\)

Ta có:

\(d:3x - 2y - 5 = 0\)\( \Rightarrow {\vec n_d} = \left( {3;\,\, - 2} \right)\)

\(d':kx - y + 1 = 0\)\( \Rightarrow {\vec n_{d'}} = \left( {k;\,\, - 1} \right)\)

Theo đề bài, ta có:

\(\cos \left( {d,\,\,d'} \right) = \cos {45^0} = \dfrac{{\left| {{{\vec n}_d}.{{\vec n}_{d'}}} \right|}}{{\left| {{{\vec n}_d}} \right|.\left| {{{\vec n}_{d'}}} \right|}}\)\( = \dfrac{{\left| {3.k + \left( { - 2} \right).\left( { - 1} \right)} \right|}}{{\sqrt {{3^2} + {{\left( { - 2} \right)}^2}} .\sqrt {{k^2} + 1} }}\)

\(\begin{array}{l} \Leftrightarrow \dfrac{{\sqrt 2 }}{2} = \dfrac{{\left| {3.k + \left( { - 2} \right).\left( { - 1} \right)} \right|}}{{\sqrt {{3^2} + {{\left( { - 2} \right)}^2}} .\sqrt {{k^2} + 1} }}\\ \Leftrightarrow \dfrac{{\sqrt 2 }}{2} = \dfrac{{\left| {3k + 2} \right|}}{{\sqrt {13\left( {{k^2} + 1} \right)} }}\\ \Leftrightarrow \dfrac{2}{4} = \dfrac{{9{k^2} + 12k + 4}}{{13{k^2} + 13}}\\ \Leftrightarrow 26{k^2} + 26 = 36{k^2} + 48k + 16\\ \Leftrightarrow 10{k^2} + 48k - 10 = 0\\ \Leftrightarrow \left( {k + 5} \right)\left( {5k - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}k + 5 = 0\\5k - 1 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}k =  - 5\\k = \dfrac{1}{5}\end{array} \right.\end{array}\)

Chọn B.