6. Let $P, A, B, C$ be four distinct points in space, and $\alpha \boldsymbol{P A}+\beta \boldsymbol{P B}+\gamma \boldsymbol{P C}=\mathbf{0}(\alpha, \beta, \gamma \in \mathbf{R})$.
Then $\alpha+\beta+\gamma=0$ and $\alpha \beta \gamma \neq 0$ is a ( ) for $A, B, C$ to be collinear.
(A) necessary and sufficient condition
(B) sufficient but not necessary condition
(C) necessary but not sufficient condition
(D) neither sufficient nor necessary condition
6.B.

Sufficiency.
From $\alpha+\beta+\gamma=0$, we get $\alpha=-\beta-\gamma$. Therefore,
$$
\begin{array}{l}
(-\beta-\gamma) \boldsymbol{P A}+\beta \boldsymbol{P} \boldsymbol{B}+\gamma \boldsymbol{P C}=\mathbf{0} \\
\Rightarrow \beta(\boldsymbol{P} \boldsymbol{B}-\boldsymbol{P A})+\gamma(\boldsymbol{P} \boldsymbol{C}-\boldsymbol{P A})=\mathbf{0} \\
\Rightarrow \beta \boldsymbol{B}+\gamma \boldsymbol{A} \boldsymbol{C}=\mathbf{0} .
\end{array}
$$

Since $\alpha \beta \gamma \neq 0$, it follows that points $A$, $B$, and $C$ are collinear.
Necessity.
When point $P$ is not on the line $B C$, by the collinearity of points $A$, $B$, and $C$, there exists a real number $x$ such that $\boldsymbol{A B}=x \boldsymbol{B C}$. Hence,
$$
\begin{array}{l}
\boldsymbol{P B}-\boldsymbol{P A}=x(\boldsymbol{P C}-\boldsymbol{P B}) \\
\Rightarrow-\boldsymbol{P A}+(1+x) \boldsymbol{P B}-x \boldsymbol{P C}=\mathbf{0} \\
\Rightarrow \frac{-1}{\alpha}=\frac{1+x}{\beta}=\frac{-x}{\gamma} \\
\Rightarrow-\frac{1}{\alpha}=\frac{1}{\beta+\gamma} \Rightarrow \alpha+\beta+\gamma=0 .
\end{array}
$$

When point $P$ is on the line $B C$, by the collinearity of points $A$, $B$, and $C$, there exists a real number $\lambda$ such that
$$
\begin{array}{l}
\boldsymbol{P A}=\lambda \boldsymbol{B C} \Rightarrow \boldsymbol{P A}=\lambda(\boldsymbol{P C}-\boldsymbol{P B}) \\
\Rightarrow \boldsymbol{P A}+\lambda \boldsymbol{P B}-\lambda \boldsymbol{P C}=\mathbf{0} \\
\Rightarrow \frac{1}{\alpha}=\frac{\lambda}{\beta}=\frac{-\lambda}{\gamma} \Rightarrow \frac{1}{\alpha}=\frac{1}{\alpha+\beta+\gamma} \\
\Rightarrow \beta+\gamma=0 .
\end{array}
$$

But $\alpha \neq 0$, thus, $\alpha+\beta+\gamma \neq 0$.