Gọi \(M\left( {0;m;0} \right) \in Oy\).

Ta có: \(\overrightarrow {AM} = \left( { - 1;m; - 2} \right)\), \(\overrightarrow {AB} = \left( {1; - 1;1} \right)\).

\( \Rightarrow \left[ {\overrightarrow {AM} ;\overrightarrow {AB} } \right] = \left( {m - 2; - 1;1 - m} \right)\).

\(\begin{array}{l} \Rightarrow {S_{MAB}} = \dfrac{1}{2}\left[ {\overrightarrow {AM} ;\overrightarrow {AB} } \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}\sqrt {{{\left( {m - 2} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( {1 - m} \right)}^2}} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}\sqrt {2{m^2} - 6m + 6} \\ \Rightarrow \dfrac{1}{2}\sqrt {2{m^2} - 6m + 6} = \dfrac{{\sqrt 6 }}{4}\\ \Leftrightarrow 2\sqrt {2{m^2} - 6m + 6} = \sqrt 6 \\ \Leftrightarrow 4\left( {2{m^2} - 6m + 6} \right) = 6\\ \Leftrightarrow 8{m^2} - 24m + 18 = 0\\ \Leftrightarrow 4{m^2} - 12m + 9 = 0\\ \Leftrightarrow {\left( {2m - 3} \right)^2} = 0\\ \Leftrightarrow m = \dfrac{3}{2}\end{array}\)

Vậy có 1 điểm \(M\) thỏa mãn yêu cầu bài toán là \(M\left( {0;\dfrac{3}{2};0} \right)\).