For a homogeneous system of linear equations to have a non-trivial solution, the determinant of its coefficient matrix must be zero.

1. Form the coefficient matrix

The system is: [ \begin{cases} 4x + \lambda y + 2z = 0 \ 2x - y + z = 0 \ \mu x + 2y + 3z = 0 \end{cases} ]

So the coefficient matrix is [ A=\begin{pmatrix} 4 & \lambda & 2\ 2 & -1 & 1\ \mu & 2 & 3 \end{pmatrix} ]

For a non-trivial solution, [ \det(A)=0 ]

2. Compute the determinant

Expand along the first row: [ \det(A)=4\begin{vmatrix}-1 & 1\2 & 3\end{vmatrix} -\lambda\begin{vmatrix}2 & 1\ \mu & 3\end{vmatrix} +2\begin{vmatrix}2 & -1\ \mu & 2\end{vmatrix} ]

Now evaluate each minor:

First minor

[ \begin{vmatrix}-1 & 1\2 & 3\end{vmatrix}=(-1)(3)-(1)(2)=-3-2=-5 ]

Second minor

[ \begin{vmatrix}2 & 1\ \mu & 3\end{vmatrix}=2\cdot 3-1\cdot \mu=6-\mu ]

Third minor

[ \begin{vmatrix}2 & -1\ \mu & 2\end{vmatrix}=2\cdot 2-(-1)\mu=4+\mu ]

So, [ \det(A)=4(-5)-\lambda(6-\mu)+2(4+\mu) ] [ = -20-\lambda(6-\mu)+8+2\mu ] [ = -12+2\mu-6\lambda+\lambda\mu ] Thus, [ \det(A)=\lambda\mu+2\mu-6\lambda-12 ]

3. Factor the expression

[ \lambda\mu+2\mu-6\lambda-12 =\mu(\lambda+2)-6(\lambda+2) ] [ =(\lambda+2)(\mu-6) ]

For non-trivial solution, [ (\lambda+2)(\mu-6)=0 ] Hence either [ \lambda=-2 \quad \text{or} \quad \mu=6 ]

4. Check the options

• A: $\mu=6,\ \lambda\in\mathbb{R}$ → True
• B: $\lambda=3,\ \mu\in\mathbb{R}$ → Not necessarily true
• C: $\mu=-6,\ \lambda\in\mathbb{R}$ → False
• D: $\lambda=2,\ \mu\in\mathbb{R}$ → False

Also note: another valid condition is $\lambda=-2$, but that is not listed.

Therefore, among the given options, the correct one is: [ \boxed{A} ]