Answer: (a) The planes are

\(\Pi_1: x+2y+3z=13\)

\(\Pi_2: x+2y+3z=-9\)

(b) A vector equation of the line is

\(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\6\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-6\\0\end{pmatrix}\)

so points on \(l\) satisfy \(\frac12x+\frac1{12}y=1\) and \(z=0\).

(c) The characteristic equation is

\(-\lambda^3+3\lambda^2+\frac74\lambda=0\)

and the eigenvalues are \(0,\ \frac72,\ -\frac12\).

(d) One suitable choice is

\(\mathbf{P}=\begin{pmatrix}3&6&-6\\-18&6&-6\\11&1&7\end{pmatrix}\)

with

\(\mathbf{D}=\begin{pmatrix}0&0&0\\0&\left(\frac72\right)^n&0\\0&0&\left(-\frac12\right)^n\end{pmatrix}\)

so that \(\mathbf{A}^n=\mathbf{PDP}^{-1}\).

(a) Since both planes are perpendicular to \(\mathbf{i}+2\mathbf{j}+3\mathbf{k}\), they both have normal vector \((1,2,3)\).

So each plane has equation of the form \(x+2y+3z=d\).

For \(\Pi_1\), the point \((1,6,0)\) lies on it, so

\(d=1+2(6)+3(0)=13\).

Hence \(\Pi_1: x+2y+3z=13\).

For \(\Pi_2\), the point \((3,-6,0)\) lies on it, so

\(d=3+2(-6)+3(0)=-9\).

Hence \(\Pi_2: x+2y+3z=-9\).

(b) The line \(l\) passes through \((1,6,0)\) and \((3,-6,0)\).

A direction vector is

\(\begin{pmatrix}3\\-6\\0\end{pmatrix}-\begin{pmatrix}1\\6\\0\end{pmatrix}=\begin{pmatrix}2\\-12\\0\end{pmatrix}\),

which can be simplified to \(\begin{pmatrix}1\\-6\\0\end{pmatrix}\).

So a vector equation of \(l\) is

\(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\6\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-6\\0\end{pmatrix}\).

This gives

\(x=1+\lambda,\quad y=6-6\lambda,\quad z=0.\)

From \(x=1+\lambda\), we have \(\lambda=x-1\).

Then

\(y=6-6(x-1)=12-6x\).

So

\(6x+y=12\).

Dividing by 12,

\(\frac12x+\frac1{12}y=1\).

Also \(z=0\).

(c) To find the characteristic equation, compute \(\det(\mathbf{A}-\lambda \mathbf{I})\):

\(\det\begin{pmatrix}1-\lambda&2&3\\1&2-\lambda&3\\\frac12&\frac1{12}&-\lambda\end{pmatrix}=0.\)

Expanding along the first row:

\((1-\lambda)\begin{vmatrix}2-\lambda&3\\\frac1{12}&-\lambda\end{vmatrix}-2\begin{vmatrix}1&3\\\frac12&-\lambda\end{vmatrix}+3\begin{vmatrix}1&2-\lambda\\\frac12&\frac1{12}\end{vmatrix}=0.\)

Now evaluate the \(2\times2\) determinants:

\(\begin{vmatrix}2-\lambda&3\\\frac1{12}&-\lambda\end{vmatrix}=(2-\lambda)(-\lambda)-3\cdot\frac1{12}=-\lambda(2-\lambda)-\frac14,\)

\(\begin{vmatrix}1&3\\\frac12&-\lambda\end{vmatrix}=-\lambda-\frac32,\)

\(\begin{vmatrix}1&2-\lambda\\\frac12&\frac1{12}\end{vmatrix}=\frac1{12}-\frac12(2-\lambda).\)

So

\((1-\lambda)\left(-\lambda(2-\lambda)-\frac14\right)-2\left(-\lambda-\frac32\right)+3\left(\frac1{12}-\frac12(2-\lambda)\right)=0.\)

Simplifying:

\(-\lambda^3+3\lambda^2-\frac74\lambda-\frac14+2\lambda+3+\frac14-3+\frac32\lambda=0,\)

so

\(-\lambda^3+3\lambda^2+\frac74\lambda=0.\)

Factorising:

\(\lambda\left(-\lambda^2+3\lambda+\frac74\right)=0.\)

Multiply the bracket by \(-4\):

\(-\lambda^2+3\lambda+\frac74=0 \iff 4\lambda^2-12\lambda-7=0.\)

Then

\(4\lambda^2-12\lambda-7=(2\lambda-7)(2\lambda+1).\)

Hence the eigenvalues are

\(\lambda=0,\quad \lambda=\frac72,\quad \lambda=-\frac12.\)

(d) We now find eigenvectors.

For \(\lambda=0\), solve \(\mathbf{A}\mathbf{v}=\mathbf{0}\):

\(\begin{pmatrix}1&2&3\\1&2&3\\\frac12&\frac1{12}&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.\)

This gives

\(x+2y+3z=0,\quad \frac12x+\frac1{12}y=0.\)

From the second equation, \(6x+y=0\), so \(y=-6x\).

Substitute into the first:

\(x+2(-6x)+3z=0 \Rightarrow -11x+3z=0 \Rightarrow z=\frac{11}{3}x.\)

Take \(x=3\). Then \(y=-18\), \(z=11\).

So one eigenvector is \(\begin{pmatrix}3\\-18\\11\end{pmatrix}\).

For \(\lambda=\frac72\), solve \((\mathbf{A}-\frac72\mathbf{I})\mathbf{v}=\mathbf{0}\).

A suitable eigenvector is \(\begin{pmatrix}6\\6\\1\end{pmatrix}\), since

\(\mathbf{A}\begin{pmatrix}6\\6\\1\end{pmatrix}=\begin{pmatrix}21\\21\\\frac72\end{pmatrix}=\frac72\begin{pmatrix}6\\6\\1\end{pmatrix}.\)

For \(\lambda=-\frac12\), solve \((\mathbf{A}+\frac12\mathbf{I})\mathbf{v}=\mathbf{0}\).

A suitable eigenvector is \(\begin{pmatrix}-6\\-6\\7\end{pmatrix}\), since

\(\mathbf{A}\begin{pmatrix}-6\\-6\\7\end{pmatrix}=\begin{pmatrix}3\\3\\-\frac72\end{pmatrix}=-\frac12\begin{pmatrix}-6\\-6\\7\end{pmatrix}.\)

So we can take

\(\mathbf{P}=\begin{pmatrix}3&6&-6\\-18&6&-6\\11&1&7\end{pmatrix}.\)

Then

\(\mathbf{A}^n=\mathbf{P}\begin{pmatrix}0&0&0\\0&\left(\frac72\right)^n&0\\0&0&\left(-\frac12\right)^n\end{pmatrix}\mathbf{P}^{-1}.\)

Hence one suitable diagonal matrix is

\(\mathbf{D}=\begin{pmatrix}0&0&0\\0&\left(\frac72\right)^n&0\\0&0&\left(-\frac12\right)^n\end{pmatrix}.\)