jmc

Since $f$ has rational coefficients, $1-2\sqrt{3}$ and $3+\sqrt{2}$ must also be roots of $f(x).$ Therefore, $f$ must be divisible by the two polynomials $$(x-(1+2\sqrt{3}))(x-(1-2\sqrt{3}))=x^2-2x-11$$and $$(x-(3-\sqrt{2}))(x-(3+\sqrt{2}))=x^2-6x+7,$$so we have $$f(x)=A(x^2-2x-11)(x^2-6x+7)$$for some constant $A.$ Setting $x=0,$ we get $$f(0)=A(-11)(7)=-77A,$$so $-77A=-154,$ and $A=2.$ Thus, $$f(x)=2(x^2-2x-11)(x^2-6x+7)$$and so $f(1)=2(-12)(2)=\boxed{-48}.$