Answer: (a) The plane \(\Pi_1\) has equation \(x+2y+z=3\).

(b) The acute angle between \(\Pi_1\) and \(\Pi_2\) is \(70.9^\circ\) (about \(1.24\) radians).

(c) A vector equation of the line of intersection is \(\mathbf{r}=\begin{pmatrix}-1\\0\\4\end{pmatrix}+t\begin{pmatrix}5\\1\\-7\end{pmatrix}\), where \(t\in\mathbb{R}\).

(a) Let \(A(2,-1,3)\), \(B(4,2,-5)\), and \(C(-1,3,-2)\). Then

\(\overrightarrow{AB}=(2,3,-8)\) and \(\overrightarrow{AC}=(-3,4,-5)\).

A normal vector to the plane is

\(\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&3&-8\\-3&4&-5\end{vmatrix}=(17,34,17).\)

This is proportional to \((1,2,1)\), so the plane has form

\(x+2y+z=d.\)

Substituting point \(A(2,-1,3)\):

\(2+2(-1)+3=3\), so \(d=3\).

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

(b) The normal vectors are \(\mathbf{n}_1=(1,2,1)\) for \(\Pi_1\) and \(\mathbf{n}_2=(3,-1,2)\) for \(\Pi_2\).

The acute angle between planes equals the acute angle between their normal vectors, so

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

Now \(\mathbf{n}_1\cdot\mathbf{n}_2=1\cdot3+2\cdot(-1)+1\cdot2=3\),

\(|\mathbf{n}_1|=\sqrt{1^2+2^2+1^2}=\sqrt6\), and \(|\mathbf{n}_2|=\sqrt{3^2+(-1)^2+2^2}=\sqrt{14}\).

Therefore

\(\cos\theta=\dfrac{3}{\sqrt6\,\sqrt{14}}=\dfrac{3}{\sqrt{84}}.\)

So \(\theta\approx 70.9^\circ\).

(c) A direction vector for the line of intersection is the cross product of the plane normals:

\(\mathbf{d}=\mathbf{n}_1\times\mathbf{n}_2=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&1\\3&-1&2\end{vmatrix}=(5,1,-7).\)

Now find a point on both planes. Set \(y=0\). Then the equations become

\(x+z=3\) and \(3x+2z=5\).

From \(x=3-z\), substitute into the second equation:

\(3(3-z)+2z=5\Rightarrow 9-z=5\Rightarrow z=4\),

so \(x=-1\).

Thus a point on the line is \((-1,0,4)\).

Hence the line of intersection is

\(\mathbf{r}=\begin{pmatrix}-1\\0\\4\end{pmatrix}+t\begin{pmatrix}5\\1\\-7\end{pmatrix},\quad t\in\mathbb{R}.\)