Actually, the matrix is somewhat arbitrary

so the key to solving this problem is to write the matrix in terms of its eigenvectors, i.e $A=S\Lambda S^{-1}$ we know we can expontiate it $e^A=S e^\Lambda S^{-1}\to det(e^A)=det(Se^\Lambda S^{-1})$ since the determinant is multiplicative $det(e^A)=det(SS^{-1})det(e^\Lambda)=det(e^\Lambda)$ remember that $\Lambda$ is a diagonal matrix, and $det(diagonal)= \prod pivots$ , so $det(e^\Lambda)=e^{\lambda_1}\times e^{\lambda_2}\times e^{\lambda_3}\to \ln(det(e^A))=\lambda_1+\lambda_2+\lambda_3$ we know the sum of the eigenvalues is the trace of the matrix, or $\lambda_1+\lambda_2+\lambda_3=x+y+z$ by cauchy swarz we have $\left(1+\frac{1}{2}+\frac{1}{3}\right)(x^2+2y^2+3z^2)\geq (x+y+z)^2 \to \boxed{\dfrac{11}{6}}\geq (x+y+z)^2=\ln^2 (det(e^A))$