Given that \(a = 0\) , show that the three planes intersect at a point.

Find the value of a such that the three planes do not meet at a point.

Given a such that the three planes do not meet at a point, find the value of \(k\) such that the planes meet in one line and find an equation of this line in the form \[\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {{x_0}} \\
  {{y_0}} \\
  {{z_0}}
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  l \\
  m \\
  n
\end{array}} \right).\]

\(\left( {\begin{array}{*{20}{c}}
  0&2&1 \\
  { - 1}&1&3 \\
  { - 2}&1&2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  3 \\
  1 \\
  k
\end{array}} \right)\)

\(\left| {\begin{array}{*{20}{c}}
  0&2&1 \\
  { - 1}&1&3 \\
  { - 2}&1&2
\end{array}} \right| = 0 - 2\left( { - 2 + 6} \right) + \left( { - 1 + 2} \right) =  - 7\)     M1A1

since determinant \( \ne 0 \Rightarrow \) unique solution to the system     R1

planes intersect in a point     AG

Note: For any method, including row reduction, leading to the explicit solution \(\left( {\frac{{6 - 5k}}{7},\frac{{10 + k}}{7},\frac{{1 - 2k}}{7}} \right)\), award M1 for an attempt at a correct method, A1 for two correct coordinates and A1 for a third correct coordinate.

[3 marks]

\(\left| {\begin{array}{*{20}{c}}
  a&2&1 \\
  { - 1}&{a + 1}&3 \\
  { - 2}&1&{a + 2}
\end{array}} \right| = a\left( {\left( {a + 1} \right)\left( {a + 2} \right) - 3} \right) - 2\left( { - 1\left( {a + 2} \right) + 6} \right) + \left( { - 1 + 2\left( {a + 1} \right)} \right)\)     M1(A1)

planes not meeting in a point \( \Rightarrow \) no unique solution i.e. determinant \(= 0\)     (M1)

\(a\left( {{a^2} + 3a - 1} \right) + \left( {2a - 8} \right) + \left( {2a + 1} \right) = 0\)

\({a^3} + 3{a^2} + 3a - 7 = 0\)     A1

\(a =1\)     A1

[5 marks]

\(\left( {\begin{array}{*{20}{c}}
  1&2&1&3 \\
  0&4&4&4 \\
  { - 2}&1&3&k
\end{array}} \right)\begin{array}{*{20}{c}}
  {{r_1} + {r_2}} \\
  {} \\
  {}
\end{array}\)     M1

\(\left( {\begin{array}{*{20}{c}}
  1&2&1&3 \\
  0&4&4&4 \\
  0&5&5&{6 + k}
\end{array}} \right)\begin{array}{*{20}{c}}
  {2{r_1} + {r_3}} \\
  {} \\
  {}
\end{array}\)     (A1)

\(\left( {\begin{array}{*{20}{c}}
  1&2&1&3 \\
  0&4&4&4 \\
  0&0&0&{4 + 4k}
\end{array}} \right)\begin{array}{*{20}{c}}
  {4{r_3} + 5{r_2}} \\
  {} \\
  {}
\end{array}\)     (A1)

for an infinite number of solutions to exist, \(4 + 4k = 0 \Rightarrow  k = - 1\)     A1

\(x + 2y + z = 3\)

\(y + z = 1\)     M1

\(\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1 \\
  1 \\
  0
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  1 \\
  { - 1} \\
  1
\end{array}} \right)\)     A1

Note: Accept methods involving elimination.

Note: Accept any equivalent form e.g. \(\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  2 \\
  0 \\
  1
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  { - 1} \\
  1 \\
  { - 1}
\end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0 \\
  2 \\
  { - 1}
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  1 \\
  { - 1} \\
  1
\end{array}} \right)\).
Award A0 if \(\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right)= \) or \(r =\) is absent.

[6 marks]

It was disappointing to see that a significant number of candidates did not appear to be well prepared for this question and made no progress at all. There were a number of schools where no candidate made any appreciable progress with the question. This was of concern since this is quite a standard problem in Mathematics HL exams. Parts (a) and (b) were intended to be answered by the use of determinants, but many candidates were not aware of this technique and used elimination. Whilst a valid method, elimination led to a long and cumbersome solution when a much more straightforward solution was available using determinants.

It was disappointing to see that a significant number of candidates did not appear to be well prepared for this question and made no progress at all. There were a number of schools where no candidate made any appreciable progress with the question. This was of concern since this is quite a standard problem in Mathematics HL exams. Parts (a) and (b) were intended to be answered by the use of determinants, but many candidates were not aware of this technique and used elimination. Whilst a valid method, elimination led to a long and cumbersome solution when a much more straightforward solution was available using determinants.

It was disappointing to see that a significant number of candidates did not appear to be well prepared for this question and made no progress at all. There were a number of schools where no candidate made any appreciable progress with the question. This was of concern since this is quite a standard problem in Mathematics HL exams. Parts (a) and (b) were intended to be answered by the use of determinants, but many candidates were not aware of this technique and used elimination. Whilst a valid method, elimination led to a long and cumbersome solution when a much more straightforward solution was available using determinants. Part (c) was also a standard question but more challenging. Very few candidates made progress on (c).