Answer: A basis for \(K_1\) is \(\{(1,1,2,-1)\}\).

A basis for \(K_2\) is \(\{(1,2,1,-1)\}\).

\(p=4\) and \(q=-4\).

To find each null space, solve \(\mathbf{M}\mathbf{x}=\mathbf{0}\).

Null space of \(T_1\)

Row reduction gives

\(\begin{pmatrix}1&-2&3&5\\0&1&4&9\\0&0&1&2\\0&0&0&0\end{pmatrix}\).

Let \(\mathbf{x}=(x_1,x_2,x_3,x_4)^T\). Then

\(x_3+2x_4=0\), so \(x_3=-2x_4\).

\(x_2+4x_3+9x_4=0\), so \(x_2=-x_4\).

\(x_1-2x_2+3x_3+5x_4=0\), so \(x_1=-x_4\).

Taking \(x_4=-t\),

\(\mathbf{x}=t(1,1,2,-1)^T\).

Thus a basis for \(K_1\) is \(\{(1,1,2,-1)\}\).

Null space of \(T_2\)

Row reduction gives

\(\begin{pmatrix}1&-2&0&-3\\0&1&0&2\\0&0&1&1\\0&0&0&0\end{pmatrix}\).

So

\(x_2+2x_4=0\), \(x_3+x_4=0\), and \(x_1-2x_2-3x_4=0\).

Taking \(x_4=-t\), we get

\(x_2=2t\), \(x_3=t\), and \(x_1=t\).

Therefore

\(\mathbf{x}=t(1,2,1,-1)^T\).

Thus a basis for \(K_2\) is \(\{(1,2,1,-1)\}\).

Now \(\mathbf{M}_1\mathbf{x}_1=\mathbf{M}_1\mathbf{a}\) implies \(\mathbf{x}_1-\mathbf{a}\in K_1\), so

\(\mathbf{x}_1=\mathbf{a}+\lambda(1,1,2,-1)^T\).

Similarly,

\(\mathbf{x}_2=\mathbf{a}+\mu(1,2,1,-1)^T\).

Hence

\(\mathbf{x}_1-\mathbf{x}_2=\lambda(1,1,2,-1)^T-\mu(1,2,1,-1)^T\)

\(=(\lambda-\mu,\lambda-2\mu,2\lambda-\mu,-\lambda+\mu)^T\).

Since this is \((p,5,7,q)^T\),

\(\lambda-2\mu=5\), and \(2\lambda-\mu=7\).

Solving gives \(\lambda=3\) and \(\mu=-1\).

Therefore

\(p=\lambda-\mu=4\), and \(q=-\lambda+\mu=-4\).