Tricky problem about quadratic forms 
Let A be symmetric matrix of order 3. Consider set
  $S=\{x\in\mathbb{R}^{3}:\; x^{T}Ax=a\}$.
1 (true). If S is unbounded for any $a\in\mathbb{R}$, then A is
  indefinite.
2 (false). If S is bounded for any $a\in\mathbb{R}$, then A is
  positive definite.
3 (false). If S contains a line for any $a\geq0$, then A is positive
  semidefinite.

Please, give me a hint how to solve this problem.
 A: Assume that $A$ is diagonal. You can assume so because the questions are independent of the basis of $\mathbb{R}^3$. So you are free to choose the coordinates to answer the questions. Do you need help assuming $A$ diagonal?
Question 1). Then (up to a change of coordinates) $S = \{(x,y,z) \in \mathbb{R}^3 : \lambda x^2 + \beta y^2 + \gamma z^2 = a \}$
Assume that $S$ is unbounded for all $a \in \mathbb{R}$. We are going to show that it is not possible that either $\lambda,\beta,\gamma \geq 0$ or $\lambda,\beta,\gamma \leq 0$. Indeed, in the first taking $a = -1$ the set $S$ is empty hence is not unbounded. In the second case take $a=1$ and again $S$ is the empty set. So at least one of $\lambda,\beta,\gamma$ is negative and the other is positive. That is to say $A$ is indefinite. So claim 1) is true.
Question 2).  As before $S = \{(x,y,z) \in \mathbb{R}^3 : \lambda x^2 + \beta y^2 + \gamma z^2 = a \}$ and assume $S$ is bounded for all $a \in \mathbb{R}$.
Observe that taking $\lambda=\beta=\gamma=-1$ we get  $S = \{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = -a \}$ which is bounded for all $a \in \mathbb{R}$ and $A$ is negative definite. Thus, claim 2) is false.
Question 3) Set $\lambda=0$,$\beta=1$ and $\gamma=-1$. Then $S = \{(x,y,z) \in \mathbb{R}^3 : y^2 - z^2 = a \}$ and $A$ is not positive semidefinite. Notice that for $a \geq 0$ the set $S$ containes the line $(t,\sqrt{a},0), t \in \mathbb{R}$. So claim 3) is false.
