(a) The direction vectors of the lines in the plane are:

\(\overrightarrow{AB} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix}, \ \overrightarrow{AC} = \begin{pmatrix} -3 \\ 0 \\ 2 \end{pmatrix}\)

The normal vector to the plane is found using the cross product:

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 4 \\ -3 & 0 & 2 \end{vmatrix} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\)

The equation of the plane is \(x - 2y + z = d\). Substituting point \(A(1, 1, 0)\) gives \(d = -1\).

(b) The perpendicular distance from \(O\) to the plane is:

\(\frac{1}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{1}{\sqrt{6}} \approx 0.408\)

(c) Let \(\overrightarrow{OP} = \begin{pmatrix} -2\lambda \\ \lambda \\ 3\lambda \end{pmatrix}\) and \(\overrightarrow{OQ} = \begin{pmatrix} 1 - 2\mu \\ 1 + \mu \\ 4\mu \end{pmatrix}\).

\(\overrightarrow{PQ} = \begin{pmatrix} 1 + \mu - \lambda \\ 4\mu - 3\lambda \end{pmatrix}\).

Using the dot product with the line direction \(\begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix}\):

\((1 - 2\mu + 2\lambda) \cdot (-2) + (1 + \mu - \lambda) \cdot 1 + (4\mu - 3\lambda) \cdot 3 = 0\)

Solving gives \(\lambda = \frac{4}{5}\).

The vector equation is \(\mathbf{r} = \frac{-4}{5} \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} + k \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}\).

(a) The vertical asymptote occurs where the denominator is zero, so \(x = 3\). For the oblique asymptote, perform polynomial long division on \(\frac{x^2 + 2x + 1}{x - 3}\):

\(y = \frac{(x-3)(x+5) + 16}{x-3} = x + 5 + \frac{16}{x-3}\)

Thus, the oblique asymptote is \(y = x + 5\).

(b) Differentiate \(y = \frac{x^2 + 2x + 1}{x - 3}\) to find turning points:

\(\frac{dy}{dx} = 1 - \frac{16}{(x-3)^2}\)

Set \(\frac{dy}{dx} = 0\):

\(1 - \frac{16}{(x-3)^2} = 0 \Rightarrow (x-3)^2 = 16\)

Solving gives \(x = -1\) and \(x = 7\). Substitute back to find \(y\)-coordinates: \((-1, 0)\) and \((7, 16)\).

(c) Sketch the curve using the asymptotes and turning points found.

(d) For \(y = \left| \frac{x^2 + 2x + 1}{x - 3} \right|\), reflect the negative parts of the curve above the x-axis. For \(y^2 = \frac{x^2 + 2x + 1}{x - 3}\), consider both positive and negative square roots, sketching both branches.