If the points with position vectors \[60\widehat i + 3\widehat j\], \[40\widehat i - 8\widehat j\], and \[a\widehat i - 52\widehat j\] are collinear then \[a\] is equal to
A. \[ - 40\]
B. \[ - 20\]
C. \[40\]
D. \[20\]

Hint: In this question, we are asked to find the value of \[a\]. For that, we first assume the given position vectors to \[A\], \[B\], and \[C\] then find the value of \[\overrightarrow {AB} \] and \[\overrightarrow {BC} \] then substitute the value of \[A\], \[B\], and \[C\] in \[\overrightarrow {AB} \] and \[\overrightarrow {BC} \] then take its determinant to find the required value.

Complete step-by-step solution:
We are given three position vectors
 \[60\widehat i + 3\widehat j\], \[40\widehat i - 8\widehat j\] and \[a\widehat i - 52\widehat j\]
Now we assume the position vector as
\[
  A = 60\widehat i + 3\widehat j\,\,...\left( 1 \right) \\
  B = 40\widehat i - 8\widehat j\,\,...\left( 2 \right) \\
  C = a\widehat i - 52\widehat j\,\,...\left( 3 \right)
 \]
Now the value of \[\overrightarrow {AB} \] is
\[\overrightarrow {AB} = \overrightarrow B - \overrightarrow A \,\,\,...\left( 4 \right)\]
Now we substitute the value of \[\overrightarrow A \] and \[\overrightarrow B \,\] from equation (1) and equation (2) in equation (4), and we get
\[
  \overrightarrow {AB} = 40\widehat i - 8\widehat j - \left( {60\widehat i + 3\widehat j} \right) \\
   = 40\widehat i - 8\widehat j - 60\widehat i - 3\widehat j \\
   = - 20\widehat i - 11\widehat j
 \]
Now the value of \[\overrightarrow {BC} \] is
\[\overrightarrow {BC} = \overrightarrow C - \overrightarrow B \,\,...\left( 5 \right)\]
Now we substitute the value of \[\overrightarrow C \,\] and \[\overrightarrow B \,\] from equation (2) and equation (3) in equation (5), and we get
\[
  \overrightarrow {BC} = a\widehat i - 52\widehat j - \left( {40\widehat i - 8\widehat j\,} \right) \\
   = a\widehat i - 52\widehat j\, - 40\widehat i + 8\widehat j\, \\
   = \left( {a - 40} \right)\widehat i\,\, - 44\widehat j\,
 \]
 Now, we know that if the vector \[A = {a_1}\widehat i + {a_2}\widehat j\] and \[B = {b_1}\widehat i + {b_2}\widehat j\] are collinear, then
 \[\left| {\begin{matrix}
  {{a_1}}&{{a_2}} \\
  {{b_1}}&{{b_2}}
\end{matrix}} \right|\, = 0\]
Now it is given that A, B, C are collinear
So, \[\left| {\begin{matrix}
  { - 20}&{ - 11} \\
  {a - 40}&{ - 44}
\end{matrix}} \right|\, = 0\]
Now we find the determinant of the above equation, we get
\[
  \left( { - 20} \right) \times \left( { - 44} \right) - \left[ {\left( {a - 40} \right) \times \left( { - 11} \right)} \right]\, = 0 \\
  880 - \left[ { - 11a + 440} \right] = 0 \\
  880 + 11a - 440 = 0 \\
  440 + 11a = 0
 \]
Further solving we get
\[
  11a = - 440 \\
  a = - \dfrac{{440}}{{11}} \\
   = - 40
 \]
Therefore, If the points with position vectors \[60\widehat i + 3\widehat j\], \[40\widehat i - 8\widehat j\], and \[a\widehat i - 52\widehat j\] are collinear then \[a\] is equal to \[ - 40\].
Hence, option(A) is correct answer

Note: Students frequently make errors when calculating the determinant and enter incorrect signs, resulting in an incorrect answer. As a result, he or she must repeat the entire process to obtain the desired result.