(a) Let line \(l_1\) pass through \(\mathbf{a} = (t,1,0)\) with direction \(\mathbf{b} = (-2,-1,0)\), and line \(l_2\) pass through \(\mathbf{c} = (0,1,t)\) with direction \(\mathbf{d} = (0,-2,1)\).

The shortest distance between two skew lines is \[ \text{distance} = \frac{\big|(\mathbf{b} \times \mathbf{d}) \cdot (\mathbf{c}-\mathbf{a})\big|} {\big|\mathbf{b} \times \mathbf{d}\big|}. \] Here \(\mathbf{b} \times \mathbf{d} = (-1,\,2,\,4)\), so \(\big|\mathbf{b} \times \mathbf{d}\big| = \sqrt{(-1)^2 + 2^2 + 4^2} = \sqrt{21}.\)

Also \(\mathbf{c} - \mathbf{a} = (0-t,\,1-1,\,t-0) = (-t,\,0,\,t)\), so \[ (\mathbf{b} \times \mathbf{d}) \cdot (\mathbf{c}-\mathbf{a}) = (-1,2,4)\cdot(-t,0,t) = t + 4t = 5t. \] Therefore \[ \text{distance} = \frac{|5t|}{\sqrt{21}}. \]

Given the shortest distance is \(\sqrt{21}\), \[ \frac{|5t|}{\sqrt{21}} = \sqrt{21} \;\Rightarrow\; |5t| = 21 \;\Rightarrow\; |t| = \frac{21}{5}. \] Since \(t\) is positive, \(t = \dfrac{21}{5}.\)

(b) The plane \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), so it contains the direction vectors of both lines: \((-2,-1,0)\) and \((0,-2,1)\), and passes through \((t,1,0) = \left(\dfrac{21}{5},1,0\right)\).

Hence an equation of \(\Pi_1\) is \[ \mathbf{r} = \frac{21}{5}\,\mathbf{i} + \mathbf{j} + \lambda(-2\mathbf{i} - \mathbf{j}) + \mu(-2\mathbf{j} + \mathbf{k}), \] where \(\lambda,\mu \in \mathbb{R}.\)

(c) The direction vector of \(l_2\) is \(\mathbf{d} = (0,-2,1)\) and a normal to \(\Pi_2\colon 5x - 6y + 7z = 0\) is \(\mathbf{n} = (5,-6,7)\).

The angle \(\alpha\) between a line and a plane satisfies \[ \sin\alpha = \frac{|\,\mathbf{d}\cdot\mathbf{n}\,|}{|\mathbf{d}|\,|\mathbf{n}|}. \] Here \(\mathbf{d}\cdot\mathbf{n} = 0\cdot5 + (-2)(-6) + 1\cdot7 = 19,\) \(|\mathbf{d}| = \sqrt{0^2+(-2)^2+1^2} = \sqrt{5},\) \(|\mathbf{n}| = \sqrt{5^2+(-6)^2+7^2} = \sqrt{110}.\)

So \[ \sin\alpha = \frac{19}{\sqrt{5}\,\sqrt{110}} = \frac{19}{\sqrt{550}}, \quad \alpha \approx 54.1^\circ. \]

(d) The acute angle between the planes \(\Pi_1\) and \(\Pi_2\) is the angle between their normals.

A normal to \(\Pi_1\) is \(\mathbf{n}_1 = \mathbf{b}\times\mathbf{d} = (-1,2,4)\), and a normal to \(\Pi_2\) is \(\mathbf{n}_2 = (5,-6,7)\).

Then \[ \cos\phi = \frac{|\,\mathbf{n}_1\cdot\mathbf{n}_2\,|} {|\mathbf{n}_1|\,|\mathbf{n}_2|} = \frac{|(-1,2,4)\cdot(5,-6,7)|} {\sqrt{21}\,\sqrt{110}} = \frac{11}{\sqrt{21\cdot110}} = \frac{11}{\sqrt{2310}}, \] so \(\phi \approx 76.8^\circ.\)