Q:

Proving Lemma 4 in Georgi's Lie Algebra in Particle Physics 2nd p 251

The lemma 4 is given in the above picture. My question is, how to verify linear dependence (20.15) for diagram (a)? I tried to extend the matrix for the simple root in wikipedia

$$
\left [\begin{matrix}
1&-1&0&0&0&0 \\
0&1&-1&0&0&0 \\
0&0&1&-1&0&0 \\
0&0&0&1&1&0 \\
-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}\\
0&0&0&1&-1&0 \\
\end{matrix}\right ]
$$
by adding 7 on top of 6 as 
$$
\left [\begin{matrix}
1&-1&0&0&0&0&0 \\
0&1&-1&0&0&0&0 \\
0&0&1&-1&0&0&0 \\
0&0&0&1&1&0&0 \\
-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}&0\\
0&0&0&1&-1&0&0 \\
x_1&x_2&x_3&x_4&x_5&x_6&x_7 
\end{matrix}\right ]
$$
since the angle between vectors 1 and 7, 4 and 7 are both zero. Therefore $x_1=x_2=0$. Since 2 and 7 are orthogonal. $x_3=0$. Since 3 and 7 are orthogonal. $x_4=0$. Since 4 and 7 are orthogonal. $x_5=0$. Since 5 and 7 are orthogonal. $x_6=0$. The vectors 6 and 7 have to be orthogonal, which is contradict with Dykin diagram.
My construction of matrix for simple roots seems to be wrong. How to verify the linear dependence correctly?

A:

Below follows the proof which Howard Georgi seems to have in mind. Let us call the root vector(s) in the Dynkin diagram $(a)$ corresponding to

the single $3$-vertex for $\vec{\gamma}$, 
the three $2$-vertices for $\vec{\beta}_1$, $\vec{\beta}_2$, $\vec{\beta}_3$, 
and the three $1$-vertices for $\vec{\alpha}_1$, $\vec{\alpha}_2$, $\vec{\alpha}_3$.

Since they are all connected through single lines, all seven root vectors have the same length, say, $\gamma$. Moreover, the pairwise angles $\theta$ are given by
$$\cos(\theta)~=~\left\{\begin{array}{rl}-\frac{1}{2}&\text{if a connecting line,} \\ \\ 0 &\text{if no connecting line.} \end{array}\right. $$
Thus we calculate
$$\left( 3\vec{\gamma}+2 \sum_{i=1}^3\vec{\beta}_i +\sum_{i=1}^3\vec{\alpha}_i \right)^2$$
$$~=~9 \vec{\gamma}^2+4 \sum_{i=1}^3\vec{\beta}_i^2 +\sum_{i=1}^3\vec{\alpha}_i^2 +12 \vec{\gamma}\cdot\sum_{i=1}^3\vec{\beta}_i + 4\sum_{i=1}^3\vec{\beta}_i\cdot \vec{\alpha}_i$$
$$~=~\gamma^2\left(9+4\cdot 3+3+ 12\cdot3\cdot(-\frac{1}{2})+ 4\cdot3\cdot(-\frac{1}{2}) \right)~=~0.$$
Hence the seven root-vectors are linearly dependent, so that the Dynkin diagram $(a)$ cannot represent a (finite-dimensional, complex) simple Lie algebra.

