= Solution
A <density matrix> has unit <trace>. The <Pauli matrices> are traceless, so $\operatorname{Tr}\hat\rho=C$ and hence $C=1$. Using $\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}$ in the <expectation value> gives
$$
\langle\sigma_j\rangle=\operatorname{Tr}(\rho\sigma_j)=r_j.
$$
Consequently the required <Bloch vector> and <density matrix> are
$$
\boxed{C=1,\qquad \mathbf r=\left(\frac12,0,0\right),\qquad
\rho=\begin{pmatrix}1/2&1/4\\1/4&1/2\end{pmatrix}.}
$$
The two <eigenvalues> are $3/4$ and $1/4$, both positive, so this is a valid <density operator>. Its <purity of a density operator> is
$$
\operatorname{Tr}\rho^2=\left(\frac34\right)^2+\left(\frac14\right)^2=\frac58<1.
$$
Therefore it is a <mixed state>. Equivalently, its <spectral decomposition> is $\rho=(3/4)|+x\rangle\langle+x|+(1/4)|-x\rangle\langle-x|$, an ensemble of the two distinct <spin one-half along an axis> states, rather than a single <pure state>.
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