Change coordinates so that the vector $g = (g_1, g_2, g_3)$ is $(1, 0, 0)$.
Now we have a quadratic $B(x) = \sum b_{ij} x_i x_j$ that's zero on the whole $x_1$-axis. The first Lemma in Chapter 3 of Samuel's Projective Geometry says (with some parts omitted)
Lemma: let $P(x_1, \ldots, x_n)$ be a homogeneous quadratic polynomial over a field $K$. Then
(a) [...]
(c) if $P(0, a_2, \ldots, a_n)$ for every $a = (a_2, \ldots, a_n) \in K^{n-1}$, then $P$ is a multiple of $x_1$.
Applying that to our situation, we see that $B(x) = x_1 H(x)$, where $H$ is necessarily homogeneous of degree $1$, i.e., a linear form, and we're done.
The proof Samuel gives is this:
Write $P(x_1, \ldots, x_n) = c_1 x_1^2 + x_1 P_1(x_2, \ldots, x_n) + P_2(x_2, \ldots, x_n)$, where $P_i$ is homogeneous of degree $i$. Now evaluating this at $(0, a_2, \ldots, a_n)$ gives $0 = 0 + 0 P_1(a_2, \ldots, a_n) + P_2(a_2, \ldots, a_n)$, hence the homogeneous quadratic polynomial $P_2$ is zero. But then
\begin{align}
P(x_1, \ldots, x_n)
&= c_1 x_1^2 + x_1 P_1(x_2, \ldots, x_n) + P_2(x_2, \ldots, x_n)\\
&= c_1 x_1^2 + x_1 P_1(x_2, \ldots, x_n) + 0\\
&= x_1 \left( c_1 x_1 + P_1(x_2, \ldots, x_n)\right).
\end{align}
[I've put words in Samuel's mouth here; he basically says "write $P$ this way; then it's clear that $P_2$ is zero."]
Best Answer
The polynomial equation for $(a, b)=(11,28)$ is given by $$ x^4+11x^3+39x^2+56x+28=(x^2 + 7x + 7)(x + 2)^2=0. $$ "How do we come up wit this layout"? By the rational root test, we find the factor $x-2$ twice, and dividing gives the factor $x^2+7x+7$. Hence the roots are $x=-2,-2,\frac{\sqrt{21}-7}{2},\frac{-\sqrt{21}-7}{2}$.
In general, the polynomial will not have any integral roots. A good example is the case $(a, b)=(1,1)$, where the polynomial $$ x^4+x^3+2x^2+2x+1 $$ is irreducible over $\Bbb Q$. It has no real root at all.