Coordinates of tilted circle. The original question is as follows:
Imagine a wire located at the intersection of $x^2+y^2+z^2=1$ and $x+y+z=0$, whose density depends on position according to $\rho({\bf x})=x^2$ per unit length. Show that the mass of the wire is $\frac{2}{3}\pi$.
I am thinking to parametrize the intersection first and do line integral over the curve. However, I can not properly write out the intersection.  Anybody has any thought on how to tackle this?

If you really want to do it the hard way, you could use the knowledge that the intersection is a circle and points on it are orthogonal to the unit vector $\hat{\eta} = (\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$. 
To find a parametrization of the circle, you need a point on the circle as a starting point.
One way to do that is pick a random vector, not in the direction
of $\pm \vec{\eta}$, project it to its components orthogonal to $\hat{\eta}$ and then
normalize it to a unit vector.
If you do this to the unit vector $\hat{x} = (1,0,0)$ in the $x$-direction, you end up
with the point $\vec{p} = (\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}} )$ lying on the circle. 
To generate the whole circle, you rotate $\vec{p}$ for some angle $\theta \in [0,2\pi)$ along the axis corresponds to $\hat{\eta}$. Let $\vec{q} = \vec{\eta} \times \vec{p} = ( 0, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} )$. The resulting locus has the
form:
$$\vec{r}(\theta) 
= \vec{p} \cos\theta + \vec{q}\sin\theta = 
\left(\frac{2\cos\theta}{\sqrt{6}}, 
-\frac{\cos\theta}{\sqrt{6}} + \frac{\sin\theta}{\sqrt{2}},
-\frac{\cos\theta}{\sqrt{6}} - \frac{\sin\theta}{\sqrt{2}} \right)$$
This is a parametrization of the circle you want. Using this, one can calculate the desired mass as
$$\int_0^{2\pi} ( \vec{r}(\theta) \cdot \hat{x} )^2 d\theta
= \int_0^{2\pi} \frac23\cos^2\theta d\theta
= \frac23\times\frac12\times 2\pi = \frac23\pi$$
