The value of $k$ for which the following system of equations possess a non-zero solution is:
$4x + ky + 2z = 0$
$kx + 4y + z = 0$
$2x + 2y + z = 0$
$\,$
$
  A) \,3 \\
  B) \,2 \\
  C) \,1 \\
  D) \,4 \\
$

Hint:Form a $3 \times 3$ determinant matrix for these linear equations by using their coefficients such as $x$ coefficient in first $y$ coefficient in second and $z$ coefficient in third. Then, find the values of $k$ for which determinant is zero.

Complete step-by-step answer:
The linear equations are
$
  4x + ky + 2z = 0 \\
  kx + 4y + z = 0 \\
  2x + 2y + z = 0 \\
 $
The condition for these linear equations to possess a non-zero solution is that the determinant formed by their coefficients is equal to zero i.e.
$\left| {\begin{array}{*{20}{c}}
  4&k&2 \\
  k&4&1 \\
  2&2&1
\end{array}} \right| = 0$
Opening the determinant
$
   \Rightarrow 4(4 - 2) - k(k - 2) + 2(2k - 8) = 0 \\
   \Rightarrow {k^2} - 6k + 8 = 0 \\
 $
Sum is $ - 6$ and product is $8$
$
   \Rightarrow {k^2} - 4k - 2k + 8 = 0 \\
   \Rightarrow k(k - 4) - 2(k - 4) = 0 \\
   \Rightarrow (k - 2)(k - 4) = 0 \\
   \Rightarrow k - 2 = 0\,\,,\,\,k - 4 = 0 \\
   \Rightarrow k = 2\,\,,\,\,k = 4 \\
 $
Hence $k$ has two values $4\,\&\,2$ for which these linear equations possess a non-zero solution.

So, the correct answer is “Option A”.

Note:The condition for which equations possess no solution is \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}\] and for for infinite many solution is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.This is correct but for two linear equations.
The condition for three equations is:
$\Delta = 0\,$ and one of ${\Delta _1},{\Delta _2},{\Delta _3}$ is zero for no solution.
$\Delta = 0\,$ and one of ${\Delta _1} = {\Delta _2} = {\Delta _3} = 0$ is zero for an infinite solution.
Where,
$\Delta = \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}} \\
  {{a_3}}&{{b_3}}&{{c_3}}
\end{array}} \right|$ , ${\Delta _1} = \left| {\begin{array}{*{20}{c}}
  {{d_1}}&{{b_1}}&{{c_1}} \\
  {{d_2}}&{{b_2}}&{{c_2}} \\
  {{d_3}}&{{b_3}}&{{c_3}}
\end{array}} \right|$ , ${\Delta _2} =\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{d_1}}&{{c_1}} \\
  {{a_2}}&{{d_2}}&{{c_2}} \\
  {{a_3}}&{{d_3}}&{{c_3}}
\end{array}} \right|$ , ${\Delta _3} = \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}}&{{d_1}} \\
  {{a_2}}&{{b_2}}&{{d_2}} \\
  {{a_3}}&{{b_3}}&{{d_3}}
\end{array}} \right|$