= Solution
The inner product of columns one and three is
$$
V_{ud}^*V_{ub}+V_{cd}^*V_{cb}+V_{td}^*V_{tb}
=A\lambda^3[(\rho-i\eta)-1+(1-\rho+i\eta)]+O(\lambda^5)=O(\lambda^5).
$$
The leading terms cancel, establishing compatibility with <unitarity> at cubic order. Direct multiplication of the printed truncation leaves $-A\lambda^5(\rho-i\eta)/2$, so it must not be treated as exactly unitary beyond its stated accuracy.
For the exact matrix, conjugate the closure relation and define
$$
a=V_{ud}V_{ub}^*,\quad b=V_{cd}V_{cb}^*,\quad c=V_{td}V_{tb}^*,\qquad a+b+c=0.
$$
Divide by $-b$ and let $z=-a/b=\bar\rho+i\bar\eta$. Then $a/(-b)=z$ and $c/(-b)=1-z$, giving a <CKM unitarity triangle> with vertices
$$
\boxed{(0,0),\qquad(1,0),\qquad(\bar\rho,\bar\eta).}
$$
At leading Wolfenstein order these are $(0,0),(1,0),(\rho,\eta)$. The barred coordinates define the exact triangle; they differ from the unbarred parameters beyond leading order.
\Image[/past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2019-unitarity-and-b-decay.png]
{title=Original normalized unitarity triangle, top-dominated neutral-B mixing box, and the leading tree topology for a common charmonium-kaon final state. The triangle coordinates are illustrative.}
For positive $\bar\eta$, define $\gamma$ at the origin, $\beta$ at $(1,0)$, and $\alpha$ at the apex. Comparing the directions of the incident complex sides gives the rephasing-invariant relations
$$
\boxed{\begin{aligned}
\alpha&=\arg\left[-\frac{V_{td}V_{tb}^*}{V_{ud}V_{ub}^*}\right],\\
\beta&=\arg\left[-\frac{V_{cd}V_{cb}^*}{V_{td}V_{tb}^*}\right],\\
\gamma&=\arg\left[-\frac{V_{ud}V_{ub}^*}{V_{cd}V_{cb}^*}\right],\qquad
\alpha+\beta+\gamma=\pi.
\end{aligned}}
$$
To leading order $\gamma=\arg(\rho+i\eta)$ and $\beta=\operatorname{atan2}(\eta,1-\rho)$; the remaining angle follows by subtraction. Each ratio cancels the unphysical common phase of the corresponding column-orthogonality terms.
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