Answer: For \(\alpha\ne 0\), a basis for \(K_1\) is \(\{(25,10,1,3)\}\).

For \(\alpha=0\), a basis for \(K_2\) is \(\{(23,8,2,0),(9,4,0,2)\}\).

Case 1: \(\alpha\ne 0\)

When \(\alpha\ne 0\), divide the fourth row by \(\alpha\). The matrix is then row-equivalent to

\(\begin{pmatrix} -2&5&3&-1\\ 0&1&-4&-2\\ 6&-14&-13&1\\ 1&1&-2&-11 \end{pmatrix}\sim \begin{pmatrix} -2&5&3&-1\\ 0&1&-4&-2\\ 0&0&3&-1\\ 0&0&0&0 \end{pmatrix}.\)

Let \((x,y,z,t)^T\in K_1\). Then the system is

\(-2x+5y+3z-t=0\),
\(y-4z-2t=0\),
\(3z-t=0\).

From \(3z-t=0\), we get \(t=3z\).

Then \(y-4z-2t=0\) gives \(y=4z+2t=10z\).

Substitute into the first equation:

\(-2x+5(10z)+3z-3z=0\), so \(-2x+50z=0\), hence \(x=25z\).

Thus

\((x,y,z,t)=z(25,10,1,3)\).

So a basis for \(K_1\) is \(\{(25,10,1,3)\}\).

Case 2: \(\alpha=0\)

Now the fourth row is zero, so the equations are

\(-2x+5y+3z-t=0\),
\(y-4z-2t=0\).

Let \(z=s\) and \(t=r\). Then

\(y=4s+2r\).

Substituting into the first equation,

\(-2x+5(4s+2r)+3s-r=0\),

so \(-2x+23s+9r=0\), and therefore

\(x=\dfrac{23s+9r}{2}\).

Choose convenient values to avoid fractions:

If \((s,r)=(2,0)\), then \((x,y,z,t)=(23,8,2,0)\).

If \((s,r)=(0,2)\), then \((x,y,z,t)=(9,4,0,2)\).

Hence

\(K_2=\operatorname{span}\{(23,8,2,0),(9,4,0,2)\}\),

so a basis for \(K_2\) is \(\{(23,8,2,0),(9,4,0,2)\}\).