$f(X,Y) = x_1y_1+6x_2y_2+3x_3y_3+2x_1y_2+2x_2y_1+3\lambda x_1y_3+3\lambda x_3y_1$.
I can write $f$ this way : $f(X,Y) =X^TAY= X^T \begin{bmatrix} 1 & 2 & 3\lambda\\ 2 & 6 & 0\\ 3\lambda & 0 & 3\\ \end{bmatrix} Y$
which means I can say that $f$ is a symmetric bilinear form
also since $1>0$, $\begin{vmatrix} 1 & 2\\ 2 & 6\\ \end{vmatrix} = 2 >0 $
therefore $A$ is postive-definite if $\begin{vmatrix} 1 & 2 & 3\lambda\\ 2 & 6 & 0\\ 3\lambda & 0 & 3\\ \end{vmatrix} >0$, right ?
$$\begin{vmatrix} 1 & 2 & 3\lambda\\ 2 & 6 & 0\\ 3\lambda & 0 & 3\\ \end{vmatrix} >0\Leftrightarrow -54\lambda +6 > 0 \Leftrightarrow \lambda <1/9$$ in conclusion $f$ defines an inner product iff $\lambda \in (-\infty,1/9)$.
I'm not 100% sure though I'm still new to inner product spaces and would like someone to confirm if I'm doing it right. thanks!