6. Given real numbers $a, b, c$ satisfy
$$
a=2 b+\sqrt{2}, a b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 \text {. }
$$

Then $\frac{b c}{a}=$ $\qquad$ .
Hint: Eliminate $a$ from the two known equations, then treat $c$ as a constant, to obtain a quadratic equation in $b$
$$
2 b^{2}+\sqrt{2} b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 .
$$

From the discriminant $\Delta \geqslant 0$, we get $c=0$.
Therefore, $\frac{b c}{a}=0$.