a) Calculate scalar product of direction of $l$ and normal to $p$

Obtain $4 \times 2 + 3 \times \left( { - 2} \right) + \left( { - 2} \right) \times 1 = 0$ and conclude accordingly

b) Substitute $\left( {\alpha ,{\text{ }}1,{\text{ }}4} \right)$ in equation of $p$ and solve for $\alpha $

Obtain $\alpha  = 4$

c) Either

Attempt use of formula for perpendicular distance using $\left( {\alpha ,{\text{ }}1,{\text{ }}4} \right)$

Obtain at least $\frac{{2\alpha  - 2 + 4 - 1}}{{\sqrt {4 + 4 + 1} }} = 6$

Obtain $\alpha  = 13$

Attempt solution of $\frac{{2\alpha  - 8}}{3} =  - 6$

Obtain $\alpha  =  - 5$

Or

Form equation of parallel plane and substitute $\left( {\alpha ,{\text{ }}1,{\text{ }}4} \right)$

Obtain $\frac{{2\alpha  + 2}}{3} - \frac{{10}}{3} = 6$

Obtain $\alpha  = 13$

Solve $\frac{{2\alpha  + 2}}{3} - \frac{{10}}{3} =  - 6$

Obtain $\alpha  =  - 5$

Or

State a vector from a pt on the plane to $\left( {\alpha ,{\text{ }}1,{\text{ }}4} \right)$ e.g.

$\left( \begin{gathered}
  \alpha  - 5 \hfill \\
  \,\,\,\,1 \hfill \\
  \,\,\,\,4 \hfill \\ 
\end{gathered}  \right)$ or $\left( \begin{gathered}
  \alpha  \hfill \\
  1 \hfill \\
   - 6 \hfill \\ 
\end{gathered}  \right)$

Calculate the component of this vector in the direction of the unit

normal and equate to $6:\frac{1}{3}\left( \begin{gathered}
  \alpha  - 5 \hfill \\
  1 \hfill \\
  4 \hfill \\ 
\end{gathered}  \right).\left( \begin{gathered}
  2 \hfill \\
   - 2 \hfill \\
  1 \hfill \\ 
\end{gathered}  \right) = 6$

Obtain $\alpha  = 13$

Solve $\frac{1}{3}\left( \begin{gathered}
  \alpha  - 5 \hfill \\
  1 \hfill \\
  4 \hfill \\ 
\end{gathered}  \right).\left( \begin{gathered}
  2 \hfill \\
   - 2 \hfill \\
  1 \hfill \\ 
\end{gathered}  \right) =  - 6$

Obtain $\alpha  =  - 5$