1. Given points $A(1,1)、B(3,2)、C(2,3)$ and line $l: y=k x(k \in \mathbf{R})$. Then the maximum value of the sum of the squares of the distances from points $A、B、C$ to line $l$ is $\qquad$
Let $A$, $B$, $C$ be three points, and the sum of the squares of their distances to the line $k x - y = 0$ is $d$. Then,
$$
\begin{aligned}
d & =\frac{(k-1)^{2}}{k^{2}+1}+\frac{(3 k-2)^{2}}{k^{2}+1}+\frac{(2 k-3)^{2}}{k^{2}+1} \\
& =\frac{14 k^{2}-26 k+14}{k^{2}+1} .
\end{aligned}
$$

Therefore, $(d-14) k^{2}+26 k+(d-14)=0$.
(1) When $d=14$, $k=0$.
(2) When $d \neq 14$, since $k \in \mathbf{R}$, we have
$$
\Delta=26^{2}-4(d-14)^{2} \geqslant 0 \text {. }
$$

Solving this, we get $1 \leqslant d \leqslant 27$. Hence, $d_{\max }=27$.