Answer: (i) \(\det\begin{pmatrix}1&2&3\\2&9&3\\1&0&4\end{pmatrix}=-1\), so the vectors are linearly independent and form a basis for \(\mathbb R^3\).

(ii) \(\mathbf d=-\mathbf a-\mathbf b+\mathbf c\).

Form the matrix whose columns are \(\mathbf a\), \(\mathbf b\), and \(\mathbf c\):

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

Then

\(\det M=1(36-0)-2(8-3)+3(0-9)=36-10-27=-1.\)

Since the determinant is non-zero, the three vectors are linearly independent. Three linearly independent vectors in \(\mathbb R^3\) form a basis, so \(\{\mathbf a,\mathbf b,\mathbf c\}\) is a basis for \(\mathbb R^3\).

Now write

\(x\mathbf a+y\mathbf b+z\mathbf c=\mathbf d.\)

This gives

\(\begin{aligned}x+2y+3z&=0,\\2x+9y+3z&=-8,\\x+4z&=3.\end{aligned}\)

From the third equation, \(x=3-4z\). Substituting into the first equation gives \(2y-z=-3\). Solving with the second equation gives

\(z=1,\qquad y=-1,\qquad x=-1.\)

Therefore

\(\mathbf d=-\mathbf a-\mathbf b+\mathbf c.\)