How to find the range of $x^{T}Ax$ 
Let $A= 
\begin{pmatrix} 
1 & -2 & -1 \\
-2 & 1 & 1 \\
-1 & 1 & 4
\end{pmatrix}, x = 
\begin{pmatrix} 
x_1\\
x_2\\
x_3
\end{pmatrix}$ with $x^{T}x = 1$, what's the range of $x^TAx$ subject to $(1,1,1)x = 0$?

We can diagonize $A$ to $\begin{pmatrix} 
-1 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 2
\end{pmatrix}$, then the question becomes finding the range of $y^T\begin{pmatrix} 
-1 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 2
\end{pmatrix}y $ with $y^Ty=1$ subject to $(\sqrt2, \frac{-2}{\sqrt 6}, \frac{-1}{\sqrt 3})y=0$, but still it seems hard to calculate, how to calculate it efficiently (by hand not computer)?

The intersection of the sphere $x^T x = 1$ with the plane $(1,1,1) \; x = 0$ is a circle.  It may be parameterized as
$$ \pmatrix{x_1\cr x_2\cr x_3\cr} = \cos(\theta) \pmatrix{1/\sqrt{2}\cr 0 \cr -1/\sqrt{2}\cr} + \sin(\theta) \pmatrix{1/\sqrt{6}\cr -2/\sqrt{6}\cr 1/\sqrt{6}\cr} $$
Compute $x^T A x$ as a function of $\theta$.  It simplifies nicely to a function of the form $a + b \cos(2\theta)+ c \sin(2\theta)$ (I'll let you find $a,b,c$), which can be written as $a + \sqrt{b^2 + c^2} \cos(2 (\theta - \theta_0))$.  So the range is $[a - \sqrt{b^2+c^2}, a + \sqrt{b^2 + c^2}]$.
