Answer
(a) Since $\overrightarrow {PQ} \times \overrightarrow {PR} = {\bf{0}}$, therefore the points $P$, $Q$, $R$ are collinear.
(b) $\overrightarrow {PQ} $ and $\overrightarrow {PR} $ form a plane containing them. The vector normal to this plane is $\overrightarrow {PQ} \times \overrightarrow {PR} = 210{\bf{i}} + 75{\bf{j}} - 45{\bf{k}}$.
(c) $\overrightarrow {PQ} $ and $\overrightarrow {PR} $ form a plane containing them. The vector normal to this plane is $\overrightarrow {PQ} \times \overrightarrow {PR} = 13{\bf{i}} - 2{\bf{j}} + 6{\bf{k}}$.
Work Step by Step
Recall from the result of Exercise 57: three points $P$, $Q$, $R$ are collinear (lie on a line) if and only if $\overrightarrow {PQ} \times \overrightarrow {PR} = {\bf{0}}$.
(a) We have $P = \left( {2,1,0} \right)$, $Q = \left( {1,5,2} \right)$, $R = \left( { - 1,13,6} \right)$. So,
$\overrightarrow {PQ} = \left( {1,5,2} \right) - \left( {2,1,0} \right) = \left( { - 1,4,2} \right),$,
$\overrightarrow {PR} = \left( { - 1,13,6} \right) - \left( {2,1,0} \right) = \left( { - 3,12,6} \right)$
Evaluate $\overrightarrow {PQ} \times \overrightarrow {PR} $:
$\overrightarrow {PQ} \times \overrightarrow {PR} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{ - 1}&4&2\\
{ - 3}&{12}&6
\end{array}} \right|$
$\overrightarrow {PQ} \times \overrightarrow {PR} = \left| {\begin{array}{*{20}{c}}
4&2\\
{12}&6
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 1}&2\\
{ - 3}&6
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
{ - 1}&4\\
{ - 3}&{12}
\end{array}} \right|{\bf{k}}$
$\overrightarrow {PQ} \times \overrightarrow {PR} = {\bf{0}}$
Since $\overrightarrow {PQ} \times \overrightarrow {PR} = {\bf{0}}$, therefore the points $P$, $Q$, $R$ are collinear.
(b) We have $P = \left( {2,1,0} \right)$, $Q = \left( { - 3,21,10} \right)$, $R = \left( {5, - 2,9} \right)$. So,
$\overrightarrow {PQ} = \left( { - 3,21,10} \right) - \left( {2,1,0} \right) = \left( { - 5,20,10} \right)$,
$\overrightarrow {PR} = \left( {5, - 2,9} \right) - \left( {2,1,0} \right) = \left( {3, - 3,9} \right)$
Evaluate $\overrightarrow {PQ} \times \overrightarrow {PR} :$:
$\overrightarrow {PQ} \times \overrightarrow {PR} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{ - 5}&{20}&{10}\\
3&{ - 3}&9
\end{array}} \right|$
$\overrightarrow {PQ} \times \overrightarrow {PR} = \left| {\begin{array}{*{20}{c}}
{20}&{10}\\
{ - 3}&9
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 5}&{10}\\
3&9
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
{ - 5}&{20}\\
3&{ - 3}
\end{array}} \right|{\bf{k}}$
$\overrightarrow {PQ} \times \overrightarrow {PR} = 210{\bf{i}} + 75{\bf{j}} - 45{\bf{k}}$
Since $\overrightarrow {PQ} \times \overrightarrow {PR} \ne {\bf{0}}$, $\overrightarrow {PQ} $ and $\overrightarrow {PR} $ form a plane containing them. The vector normal to this plane is $\overrightarrow {PQ} \times \overrightarrow {PR} = 210{\bf{i}} + 75{\bf{j}} - 45{\bf{k}}$.
(c) We have $P = \left( {1,1,0} \right)$, $Q = \left( {1, - 2, - 1} \right)$, $R = \left( {3,2, - 4} \right)$. So,
$\overrightarrow {PQ} = \left( {1, - 2, - 1} \right) - \left( {1,1,0} \right) = \left( {0, - 3, - 1} \right),$,
$\overrightarrow {PR} = \left( {3,2, - 4} \right) - \left( {1,1,0} \right) = \left( {2,1, - 4} \right)$
Evaluate $\overrightarrow {PQ} \times \overrightarrow {PR} $:
$\overrightarrow {PQ} \times \overrightarrow {PR} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
0&{ - 3}&{ - 1}\\
2&1&{ - 4}
\end{array}} \right|$
$\overrightarrow {PQ} \times \overrightarrow {PR} = \left| {\begin{array}{*{20}{c}}
{ - 3}&{ - 1}\\
1&{ - 4}
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
0&{ - 1}\\
2&{ - 4}
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
0&{ - 3}\\
2&1
\end{array}} \right|{\bf{k}}$
$\overrightarrow {PQ} \times \overrightarrow {PR} = 13{\bf{i}} - 2{\bf{j}} + 6{\bf{k}}$
Since $\overrightarrow {PQ} \times \overrightarrow {PR} \ne {\bf{0}}$, $\overrightarrow {PQ} $ and $\overrightarrow {PR} $ form a plane containing them. The vector normal to this plane is $\overrightarrow {PQ} \times \overrightarrow {PR} = 13{\bf{i}} - 2{\bf{j}} + 6{\bf{k}}$.
