3. Given a quadratic equation in $x$
$$
x^{2}-(a+b+c) x+(a b+b c+c a)=0
$$

has two equal real roots, where $a, b, c$ are distinct positive numbers. Then the triangle formed by connecting three segments of lengths $\sqrt[4]{a}, \sqrt[4]{b}, \sqrt[4]{c}$ in sequence ( ).
(A) is an acute triangle
(B) is a right triangle
(C) is an obtuse triangle
(D) such a triangle does not exist
3. B.

Notice,
$$
\Delta=0 \Rightarrow a^{2}+b^{2}+c^{2}-2 a b-2 b c-2 c a=0 \text {. }
$$

Given \( a, b, c > 0 \), factorizing the above equation yields
$$
\begin{array}{l}
-(\sqrt{a}+\sqrt{b}+\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c}) . \\
(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})=0 .
\end{array}
$$

Assume \( a > b > c \). Then
$$
\sqrt{a}=\sqrt{b}+\sqrt{c} \Rightarrow(\sqrt[4]{a})^{2}=(\sqrt[4]{b})^{2}+(\sqrt[4]{c})^{2} .
$$