Answer: Cartesian equation of \(\Pi_{1}\): \(x+3y-z=12\).

Acute angle between \(\Pi_{1}\) and \(\Pi_{2}\): \(75.7^\circ\) (approximately).

Line of intersection: \(\mathbf{r}=\begin{pmatrix}6\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\-7\end{pmatrix}\).

The plane \(\Pi_1\) is given in vector form by \(\mathbf{r}=\begin{pmatrix}2\\3\\-1\end{pmatrix}+s\begin{pmatrix}1\\0\\1\end{pmatrix}+t\begin{pmatrix}1\\-1\\-2\end{pmatrix}\). Its direction vectors are \(\begin{pmatrix}1\\0\\1\end{pmatrix}\) and \(\begin{pmatrix}1\\-1\\-2\end{pmatrix}\).

To find a normal vector, take the cross product:

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

So a normal to \(\Pi_1\) is \(\mathbf{n}_1=(1,3,-1)\). Using the point \((2,3,-1)\) on the plane,

\(1(x-2)+3(y-3)-1(z+1)=0\).

Hence \(x+3y-z=12\).

For \(\Pi_2\), the equation \(2x-y+z=10\) shows that a normal vector is \(\mathbf{n}_2=(2,-1,1)\).

The acute angle between the planes is the acute angle between their normals, so

\(\cos\theta=\dfrac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}\).

Now \(\mathbf{n}_1\cdot\mathbf{n}_2=1\cdot2+3\cdot(-1)+(-1)\cdot1=2-3-1=-2\), so \(|\mathbf{n}_1\cdot\mathbf{n}_2|=2\).

Also \(|\mathbf{n}_1|=\sqrt{1^2+3^2+(-1)^2}=\sqrt{11}\) and \(|\mathbf{n}_2|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}\).

Therefore \(\cos\theta=\dfrac{2}{\sqrt{66}}\), so \(\theta\approx 75.7^\circ\).

For the line of intersection, its direction vector is perpendicular to both normals, so use the cross product:

\(\mathbf{d}=\mathbf{n}_1\times\mathbf{n}_2=\begin{pmatrix}1\\3\\-1\end{pmatrix}\times\begin{pmatrix}2\\-1\\1\end{pmatrix}=\begin{pmatrix}2\\-3\\-7\end{pmatrix}\).

Now find a point common to both planes. A convenient choice is to set \(z=0\). Then the plane equations become

\(x+3y=12\) and \(2x-y=10\).

Solving: from \(2x-y=10\), \(y=2x-10\). Substitute into \(x+3y=12\):

\(x+3(2x-10)=12\)

\(7x=42\)

\(x=6\), so \(y=2\).

Thus \((6,2,0)\) lies on both planes.

So the line of intersection is

\(\mathbf{r}=\begin{pmatrix}6\\2\\0\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\-7\end{pmatrix}\).