2. Given three points $A(-2,1)$, $B(-3,-2)$, $C(-1,-3)$, and a moving line $l: y=kx$. When the sum of the squares of the distances from points $A$, $B$, and $C$ to the line $l$ is minimized, which of the following conclusions is correct? ( ).
(A) Point $A$ lies on the line $l$
(B) Point $B$ lies on the line $l$
(C) Point $C$ lies on the line $l$
(D) Points $A$, $B$, and $C$ do not lie on the line $l$
2. (D).

Points $A, B, C$ to the line $l: k x-y=0$ have a sum of the squares of the distances as
$$
\begin{aligned}
d & =\frac{(-2 k-1)^{2}+(-3 k+2)^{2}+(-k+3)^{2}}{k^{2}+1} \\
& =\frac{14 k^{2}-14 k+14}{k^{2}+1}=14-\frac{14 k}{k^{2}+1} .
\end{aligned}
$$

To minimize $d$, it is clear that $k>0$. At this time, $d \geqslant 14-\frac{14 k}{2 k}=7$, when and only when $k^{2}=1$, i.e., $k=1$.

Therefore, when $k=1$, $d$ reaches its minimum value. At this point, $l$ is $y=x$, and it is evident that points $A, B, C$ are not on the line $l$.