normally, i find eigenvectors by finding the eigenvalues first. this involves finding det(A - xI) for your matrix A.

in this particular case, the 0's actually make it easier to take the determinant, and det(A - xI) =

-(7/10 + x)(3/10 + x)(1/10 + x). setting this equal to 0 gives the 3 eigenvalues: -7/10, -3/10 and -1/10.

so let's see what we can come up with for the eigenvalue -7/10. we know that for an eigenvector v for the eigenvalue -7/10

Av + (7/10)v = 0. in matrix form:

[..0.. ..0.. .....0][v1].....[0]

[7/10 4/10 ....0][v2]....[0]

[..0.. 3/10 6/10][v3] = [0]

since the first row is all 0, it doesn't tell us anything, the first entry of (A +(7/10)I)v will be 0 no matter what v is.

but from the 2nd row we see that (7/10)v1 + (4/10)v2 = 0, so v1 = (-4/7)v2.

and from the 3rd row, we have (3/10)v2 + (6/10)v3 = 0, so v3 = (-1/2)v2.

we can thus pick any value we like for v2, and this will determine v1 and v3. to avoid fractions, let's pick v2 = -14.

then v1 = (-4/7)v2 = (-4/7)(-14) = 8, and v3 = (-1/2)(-14) = 7, so v = (8,-14,7).

( (-8,14,-7) and (8/7,-2,1) would also be perfectly acceptable eigenvectors).

the other two eigenvectors can be calculated in the same way.