Let $0,1,2$ is eigenvalues and $x_1,x_2,x_3$ eigenvectors of matrix $A$:
a) Describe subspace $\ker(A),\operatorname{Im}(A),R(A^{T})$ in terms of $x1,x2,x3$
b) Find all solution $Ax=2x_2-3x_3$ in term of $x_1,x_2,x_3$. What you can say about system $Ax=x_1+x_3$?
c) Check is matrix $A$ an orthogonal matrix?
If we have 0 for eigenvalue than $\dim\ker(A)=1$, and $\dim\operatorname{Im}(A)=2$ since $ket(A)\oplus \operatorname{Im}(A^{T})$ then $\dim\operatorname{Im}(A^{T})=2$, I do not know what to tell more. For b) if we know that $Ax_1=0,Ax_2=1x_2,Ax_3=2x_3$ then we can write $Ax=2Ax_2-\frac{3}{2}Ax_3$ then $x=2x_2-\frac{3}{2}x_3$ so all solution is $L(x_2,x_3)$, other part I only see this $A^{2}x=Ax_1+Ax_3=Ax_3$ so $Ax=x_3$ so $x=\frac{1}{2}x_3$.For c) I think it is not because we have $(Ax_1,Ax_2)=0$ and $(Ax_1,Ax_3)=0$, but $(Ax_2,Ax_3)\not=0$. What is your opinion?
a) Do you know about the spectral decomposition: $A = S\Lambda S^{-1}$
b) For you to deduce $x = 2x_2 - \frac 3{2}x_3$ from the equation $Ax = 2Ax_2 - \frac 3{2}Ax_3$, $A$ must have an inverse. But A is singular.
c) $Ax_1$ is a zero vector. You cannot dot product zero vector with a column vector to find if they are orthogonal because every vector is orthogonal to zero vector. You need to find the basis for column space and check if it can be orthogonalized.