7. Given $a, b, c \geqslant 0, t \geqslant 1$, satisfying
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{2}, \\
\sqrt{a+\frac{1}{2}(b-c)^{2}}+\sqrt{b}+\sqrt{c}=\frac{\sqrt{6 t}}{2} .
\end{array}\right.
$$

Then $a^{2 t}+b^{2 t}+c^{2 t}=$ $\qquad$ .
7. $\frac{1}{12}$.

Assume $b \geqslant c$, let $\sqrt{2 b}=x+y, \sqrt{2 c}=x-y$.
Then $b-c=2 x y, 2 a=1-2 x^{2}-2 y^{2} \geqslant 0$.
Thus, $0 \leqslant x \leqslant \frac{\sqrt{2}}{2}$.
$$
\begin{array}{l}
\text { Hence } \sqrt{2 a+(b-c)^{2}}+\sqrt{2 b}+\sqrt{2 c} \\
=\sqrt{\left(1-2 x^{2}\right)\left(1-2 y^{2}\right)}+x+x \\
\leqslant \sqrt{1-2 x^{2}}+x+x \\
\leqslant \sqrt{\left(1-2 x^{2}+x^{2}+x^{2}\right)\left(1^{2}+1^{2}+1^{2}\right)} \\
=\sqrt{3} .
\end{array}
$$

From the given equation, the above inequality holds with equality, i.e.,
$$
\begin{array}{l}
y^{2}=0,1-2 x^{2}=x^{2} \Rightarrow x=\frac{\sqrt{3}}{3}, y=0 \\
\Rightarrow a=b=c=\frac{1}{6}, t=1 \\
\Rightarrow a^{2 t}+b^{2 t}+c^{2 t}=\frac{1}{12} .
\end{array}
$$