This is partly a physics question, but here it is:
I have an inertia tensor, I have found the eigevalues as 1, 1 and 4.
To find the eigenvectors I have found the simultaneous equations
\begin{align} 2a + b + c &= a; \\ a + 2b + c &= b; \\ a + b + 2c &= c; \end{align}
For eigenvalue=1, I get the relations $a-b = a-b$; $b-c= b-c$, which obviously don't specify a particular value of $a$ or $b$.
For eigenvalue=1 I can find multiple eigenvectors which satisfy the eigenvalue equation, in addition to the ones in the past paper answers which this question is from.
What is the mathematical reason for this?
Please review your knowledge for eigenvector computations. Your system is the eigenvector system for the double eigenvalue 1. Unsurprisingly, since the matrix is symmetric, the eigen subspace is two-dimensional, i.e., each of the equations is equivalent to $$a+b+c=0.$$ For instance $(a,b,c)=(1,0,-1)$ and $(a,b,c)=(-1,2,-1)$ are two orthogonal eigenvectors.
For the eigenvalue 4, the system is
\begin{align} 2a + b + c &= 4a; \\ a + 2b + c &= 4b; \\ a + b + 2c &= 4c; \end{align}
which can be reformulated as $b-a=a-c=c-b$, so the only eigenvector, up to multiples, is $(a,b,c)=(1,1,1)$.