Determine whether the following quadratic forms are isotropic over $\mathbb{Q}$. 
*

*$q(x, y, z) = 5x^2-y^2-11z^2$

*$q(x, y, z) = 3x^2-y^2+22z^2$
So for part (1), I apply Hasse-Minkowski and check over the $\mathbb{Q}_p$. It's trivial to see for $\mathbb{R}$. We only need to check for $\mathbb{Q}_{13}$ and $\mathbb{Q}_5$ right?. Now mod 5, the equation becomes $-y^2-z^2 = 0 \mod 5$  and thus there is a solution since -1 is square mod 5. Checking mod 11, this becomes $5x^2-y^2 = 0 \mod 11$. Now enough to check if 5 is square mod 11. But it's clear that it is by Legendre. Thus there is a solution and the form isotropic over $\mathbb{Q}$.
Part(2) is similar. We check mod 3, $-y^2+z^2 = 0 \mod 3$ and clearly there is a solution. For mod 11, we have that $3x^2-y^2 = 0 \mod 11$. Now 3 is a square mod 11 thus there is a solution. Thus it's isotropic over $\mathbb{Q}$.
Note that for other $p$, the coefficients are units thus must be isotropic. 
Am I right? 
 A: next day: those recipes below give all primitive. Two small solutions to $5x^2 = y^2 + 11 z^2$ are $(2,3,1)$ and $(3,1,2).$ Two small solutions to $y^2 = 3x^2 + 22 z^2$ are $(1,5,1)$ and $(3,7,1).$ 
We get infinitely many primitive solutions for the first one taking 
$$ x = 5 u^2 + 6 uv + 4 v^2 \; \; , \; \;   y = 9 u^2 + 2 uv -6 v^2 \; \; , \; \;  z = 2 u^2 + 6 uv + 2 v^2 \; \; , \; \;  $$
and the genuinely different (even considering absolute values)
$$ x = 6 u^2 + 2 uv + 2 v^2 \; \; , \; \;   y = 13 u^2 + 8 uv -3 v^2 \; \; , \; \;  z =  u^2 -4 uv - v^2 \; \; . \; \;  $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Second one
$$ x = 3 u^2 + 8 uv -2 v^2 \; \; , \; \;   y = 7 u^2 + 4 uv + 10 v^2 \; \; , \; \;  z =  u^2 - 2 uv - 2 v^2 \; \; , \; \;  $$
and the different
$$ x =  u^2 + 8 uv -6 v^2 \; \; , \; \;   y = 5 u^2 - 4 uv + 14 v^2 \; \; , \; \;  z =  u^2 - 2 uv - 2 v^2 \; \; . \; \;  $$
In both cases you may take absolute values of $x,y,z$ as there are no mixed terms.
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= 
The example I like to show is solving
$$ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0,  $$ four "recipes,"
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
37 u^2 + 51 uv + 8 v^2 \\
8 u^2 -35 uv -6 v^2 \\
-6 u^2 +  23 uv + 37 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
32 u^2 + 61 uv + 18 v^2 \\
18 u^2 -25 uv -11 v^2 \\
-11 u^2 +  3 uv + 32 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
38 u^2 + 45 uv + 4 v^2 \\
4 u^2 -37 uv -3 v^2 \\
-3 u^2 +  31 uv + 38 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
29 u^2 + 63 uv + 22 v^2 \\
22 u^2 -19 uv -12 v^2 \\
-12 u^2 -5 uv + 29 v^2
\end{array}
\right)
$$
For all four recipes,
$$ x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2  \right)^2  $$
