Answer: (i) The plane \(\Pi_1\) has two direction vectors \(\langle 4,-4,8\rangle\) and \(\langle 1,2,3\rangle\). A normal vector is therefore \(\langle 7,1,-3\rangle\), so the cartesian equation is \(7x+y-3z=6\).

(ii) The plane \(\Pi_2\) has direction vectors \(\langle 1,-2,-1\rangle\) and \(\langle 2,3,-1\rangle\). A normal vector is \(\langle 5,-1,7\rangle\), so the cartesian equation is \(5x-y+7z=20\).

(iii) The acute angle between the planes is \(78.7^\circ\) (approximately).

(i) Since \(\Pi_1\) passes through \(A(1,2,1)\) and \(B(5,-2,9)\), one direction vector in the plane is

\(\overrightarrow{AB}=(5-1,-2-2,9-1)=(4,-4,8)=4(1,-1,2).\)

The plane is also parallel to \(\mathbf{i}+2\mathbf{j}+3\mathbf{k}\), so a second direction vector is \(\langle 1,2,3\rangle\).

A normal vector is found by taking the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 1 & 2 & 3 \end{vmatrix}=\langle -7,-1,3\rangle.\)

So a normal vector may be taken as \(\langle 7,1,-3\rangle\). Hence the equation of the plane has form

\(7x+y-3z=d.\)

Substitute the point \((1,2,1)\):

\(7(1)+2-3(1)=6,\)

so \(d=6\). Therefore

\(\Pi_1: 7x+y-3z=6.\)

(ii) The plane \(\Pi_2\) contains the two lines with direction vectors

\(\langle 1,-2,-1\rangle\) and \(\langle 2,3,-1\rangle\).

A normal vector is their cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -1 \\ 2 & 3 & -1 \end{vmatrix}=\langle 5,-1,7\rangle.\)

So the plane has equation

\(5x-y+7z=d.\)

The point \((2,-3,1)\) lies on both lines, so it lies in the plane. Substitute it:

\(5(2)-(-3)+7(1)=10+3+7=20.\)

Hence

\(\Pi_2: 5x-y+7z=20.\)

(iii) The angle between two planes is the angle between their normal vectors. Using normals \(\mathbf{n}_1=\langle 7,1,-3\rangle\) and \(\mathbf{n}_2=\langle 5,-1,7\rangle\),

\(\cos\theta=\left|\frac{\mathbf{n}_1\cdot \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|}\right|.\)

Now

\(\mathbf{n}_1\cdot \mathbf{n}_2=7\cdot 5+1\cdot (-1)+(-3)\cdot 7=35-1-21=13,\)

and

\(|\mathbf{n}_1|=\sqrt{7^2+1^2+(-3)^2}=\sqrt{59},\\qquad |\mathbf{n}_2|=\sqrt{5^2+(-1)^2+7^2}=\sqrt{75}.\)

So

\(\cos\theta=\frac{13}{\sqrt{59}\sqrt{75}}.\)

Therefore the acute angle is

\(\theta\approx 78.7^\circ.\)