6. Let the vectors be
$$
a_{1}=(1,5), a_{2}=(4,-1), a_{3}=(2,1) \text {, }
$$
$\lambda_{1} 、 \lambda_{2} 、 \lambda_{3}$ are non-negative real numbers, $\lambda_{1}+\frac{\lambda_{2}}{2}+\frac{\lambda_{3}}{3}=1$. Then the minimum value of $\left|\lambda_{1} a_{1}+\lambda_{2} a_{2}+\lambda_{3} a_{3}\right|$ is $\qquad$
$6.3 \sqrt{2}$.
$$
\begin{array}{l}
\text { Let } b_{1}=a_{1}=(1,5), b_{2}=2 a_{2}=(8,-2), \\
b_{3}=3 a_{3}=(6,3), p_{1}=\lambda_{1}, p_{2}=\frac{\lambda_{2}}{2}, p_{3}=\frac{\lambda_{3}}{3} . \\
\text { Then } \lambda_{1} a_{1}+\lambda_{2} a_{2}+\lambda_{3} a_{3}=p_{1} b_{1}+p_{2} b_{2}+p_{3} b_{3} .
\end{array}
$$

Thus, the problem is transformed into finding the minimum value of $\left|p_{1} \boldsymbol{b}_{1}+p_{2} \boldsymbol{b}_{2}+p_{3} \boldsymbol{b}_{3}\right|$ under the conditions that $p_{1} 、 p_{2} 、 p_{3}$ are non-negative real numbers, and $p_{1}+p_{2}+p_{3}=1$.
$$
\begin{array}{l}
\text { Let } O(0,0), A(1,5), B(8,-2), C(6,3), \\
\overrightarrow{O P}=p_{1} \boldsymbol{b}_{1}+p_{2} \boldsymbol{b}_{2}+p_{3} \boldsymbol{b}_{3} \text {. }
\end{array}
$$

Then the conditions that $p_{1} 、 p_{2} 、 p_{3}$ are non-negative real numbers and $p_{1}+p_{2}+p_{3}=1$ correspond to $P$ being a point inside (including the boundary) of $\triangle A B C$.

From the graph, we know that when $P$ is the point $(3,3)$ on the line segment $A B$, $O P \perp A B$, at this time,
$$
|\overrightarrow{O P}|=\left|p_{1} \boldsymbol{b}_{1}+p_{2} \boldsymbol{b}_{2}+p_{3} \boldsymbol{b}_{3}\right|
$$

achieves its minimum value $3 \sqrt{2}$.