Isotropy over $p$-adic numbers Over what $p$-adic fields $\mathbb{Q}_p$ is the form $\langle3, 7, -15\rangle$ isotropic?
 A: Given your questions, you absolutely must purchase the Cassels book, and quickly, otherwise you are completely screwed.
http://www.amazon.com/Rational-Quadratic-Forms-Dover-Mathematics/dp/0486466701
We find tables of the Hilbert Norm Residue Symbol $(a,b)_p$ on pages 43 and 44, one for $\mathbb Q_p$ with odd prime $p,$ a bigger table for $\mathbb Q_2,$ then a smaller one for $\mathbb Q_\infty.$ 
On page 55, for a quadratic form 
$$  f(x_1, \ldots, x_n) = a_1 x_1^2 + \ldots + a_n x_n^2, \; \; a_j \in \mathbb Q_p^\ast,$$
he gives the definition 
$$  c_p(f) = \prod_{i < j} (a_i, a_j)_p  $$
which is his version of the Hasse-Minkowski Invariant.
On page 59, we have Lemma 2.5, which says a (nondegenerate) ternary is isotropic in $\mathbb Q_p$ if and only if
$$ c_p(f) = (-1, - \det(f))_p$$
Then we have the calculations. This is intricate and prone to error, but not conceptually difficult at this stage. 
$$ 
   \begin{array}{ccccccc}
 p & (3,7)_p & (3,-15)_p & (7,-15)_p & c_p & (-1,315)_p & \mbox{comment}  \\
 2 & -1 & 1 & 1 & -1 & -1 &  315 \equiv 3 \pmod 8  \\
 3 & 1 & -1 & 1 & -1 & 1 & 35 = \frac{315}{9} \equiv -1 \pmod 3  \\
 5 & 1 & -1 & -1 & 1 & 1 &  63 = \frac{315}{5} \equiv 3 \pmod 5 \\
 7 & -1 & 1 & -1 & 1 & -1 &  45 = \frac{315}{7} \equiv 3 \pmod 7  \\
 \infty & 1 & 1 & 1 & 1 & 1 & \mbox{indefinite} 
\end{array} 
   $$
Some authors define their version of the Hasse-Minkowski Invariant as the quantity that Cassels would call 
$$   c_p(f) \; \cdot \;   (-1, - \det(f))_p,$$ in which case you get isotropy if and only if the number is 1.
A: Everything is in Cassels, Rational Quadratic Forms, pages 41-44, then pages 55-59. In particular Lemma 2.5 on page 59. Your forms is isotropic at the primes $\{2,5,\infty\}.$ It is anisotropic at primes $\{3,7\}.$ And isotropic at everything else. 
To give the simplest looking version, what happens if $$  3 x^2 + 7 y^2 - 15 z^2 \equiv 0 \pmod {49}?$$
Well, $$  3 x^2  - 15 z^2 \equiv 0 \pmod {7},$$
 $$   x^2  - 5 z^2 \equiv 0 \pmod {7},$$
$$   x^2  \equiv 5 z^2  \pmod {7}.$$
If we assume $z$ is not divisible by 7, it has a multiplicative inverse $\pmod 7,$ so
$$  \frac{ x^2}{z^2}  \equiv 5   \pmod {7}$$ and
$$  \left(\frac{ x}{z}\right)^2  \equiv 5   \pmod {7}.$$ 
However, 5 is a quadratic nonresidue $\pmod 7,$ written $(5|7)= -1.$ 
So the equation is impossible, contradicting the assumption, and, in fact, $z$ is divisible by 7. From $   x^2  \equiv 5 z^2  \pmod {7}$ we find that $x$ is also divisible by 7. So, actually, $$   x^2  - 5 z^2 \equiv 0 \pmod {49}.$$
It follows from $  3 x^2 + 7 y^2 - 15 z^2 \equiv 0 \pmod {49}$ that $7 y^2  \equiv 0 \pmod {49}$ which tells us that $ y^2  \equiv 0 \pmod {7}$ and $ y  \equiv 0 \pmod {7}.$ In sum, $  3 x^2 + 7 y^2 - 15 z^2 \equiv 0 \pmod {49}$ implies that $(x,y,z)$ are all divisible by 7. It follows that we can divide out 7 from each of the variables and 49 from whatever the result was. Or, we cannot express a number divisible by an arbitrarily high power of 7 unless $(x,y,z)$ are all divisible by 7 to roughly half that exponent.
Or, put another way, the only solution to  $  3 x^2 + 7 y^2 - 15 z^2 = 0 $ in
$\mathbb Q_7$ is $(x=0, y=0, z=0).$ 
It is a worthwhile exercise for you to confirm things about the prime 3, as above. Also plow through Lemma 2.5 for each prime. 
Finally, from Lemma 2.4 on page 58, $7 y^2 - 15 z^2$ is isotropic in $\mathbb Q_2,$ and $ 3 x^2 + 7 y^2$ is isotropic in $\mathbb Q_5.$
