= Solution
For the <Cholesky decomposition> $A=LL^T$, choose $L$ lower triangular with positive diagonal whenever $A$ is <positive-definite>. Successive entries are
$$
l_{11}=\sqrt2,\quad l_{21}=-2\sqrt2,\quad l_{31}=\sqrt2,\quad
l_{22}=\sqrt{\lambda+2}.
$$
The $(3,2)$ equation gives $l_{31}l_{21}+l_{32}l_{22}=2+3\lambda$, so $l_{32}=3\sqrt{\lambda+2}$. Finally,
$$
l_{33}^2=23+9\lambda-l_{31}^2-l_{32}^2=3.
$$
Thus for $\lambda>-2$,
$$
\boxed{L=\begin{pmatrix}
\sqrt2&0&0\\
-2\sqrt2&\sqrt{\lambda+2}&0\\
\sqrt2&3\sqrt{\lambda+2}&\sqrt3
\end{pmatrix}}.
$$
It is nonsingular, and for every nonzero vector $v$, $v^TAv=|L^Tv|^2>0$.
Necessity also follows from the leading principal determinants $2$, $2(\lambda+2)$ and $6(\lambda+2)$, or directly from the second elimination pivot. Hence
$$
\boxed{A\text{ is positive definite exactly when }\lambda>-2}.
$$
At $\lambda=-2$ the displayed limiting factor still satisfies $A=LL^T$, but has rank two and zero second diagonal entry: $A$ is <positive semidefinite>, not <positive-definite>. For $\lambda<-2$, taking $v=(2,1,0)^T$ gives $v^TAv=\lambda+2<0$, so no real factor $LL^T$ exists.
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