Roots of the characteristic polynomial are eigenvalues of the matrix and the characteristic polynomial of matrix D is
Char(D)= det(D - xI), where I is the identity matrix of size 3.
Write this determinant using the Laplace expansion by the first column, so you get
det( D-xI) = (4-x) [ (e-x)(6-x) + 2f ].
Note that this the polynomial is of degree 3. As all three eigenvalues are identical (the assumption), this polynomial has only one root with multiplicity 3.
Since det( D-xI) = (4-x) [ (e-x)(6-x) + 2f ] = 0, so 4 must be the only solution. So, 4 is also the solution of the part
[ (e-x)(6-x) + 2f ] . Hence
(e-4)(6-4)+2f = 0 ⇒ f = 4 - e.
As 4 is the only solution, the discriminate of the polynomial (e-x)(6-x) + 2f = x2 - (e+6) x + 6e+2f must be 0.
Δ = (e+6)2 - 24e - 8f = 0.
Replacing f = 4 - e in the equation above we get
Δ = (e+6)2 - 24e - 8(4-e) = (e+6)2 - 16e - 32 = e2 - 4e + 4 = (e-2)2 = 0.
Hence e=2 and so f=2.
Now you have the matrix D, and you know the eigenvalues is x=4, find the associated eigenvector by looking at the null space of the matrix D-4I, where I is the identity matrix.
You may look at the following calculator and it shows step by step solution