We need the values of $\lambda$ for which the given homogeneous-looking system has a non-trivial solution in $x,y,z$.

1. Rewrite the system in standard linear form

Given: [ \begin{aligned} x-2y-2z &= \lambda x \ x+2y+z &= \lambda y \ -x-y &= \lambda z \end{aligned} ] Bring all terms to the left: [ \begin{aligned} (1-\lambda)x-2y-2z &= 0 \ x+(2-\lambda)y+z &= 0 \ -x-y-\lambda z &= 0 \end{aligned} ] So the coefficient matrix is [ A=\begin{pmatrix} 1-\lambda & -2 & -2\ 1 & 2-\lambda & 1\ -1 & -1 & -\lambda \end{pmatrix} ]

For a homogeneous system to have a non-trivial solution, we need [ \det(A)=0. ]

2. Compute the determinant

[ \det(A)= \begin{vmatrix} 1-\lambda & -2 & -2\ 1 & 2-\lambda & 1\ -1 & -1 & -\lambda \end{vmatrix} ] Expand along the first row: [ \det(A)=(1-\lambda) \begin{vmatrix} 2-\lambda & 1\ -1 & -\lambda \end{vmatrix} -(-2) \begin{vmatrix} 1 & 1\ -1 & -\lambda \end{vmatrix} +(-2) \begin{vmatrix} 1 & 2-\lambda\ -1 & -1 \end{vmatrix} ] Now evaluate each minor:

First minor

[ \begin{vmatrix} 2-\lambda & 1\ -1 & -\lambda \end{vmatrix} =(2-\lambda)(-\lambda)-1(-1) =-2\lambda+\lambda^2+1 =\lambda^2-2\lambda+1=(\lambda-1)^2 ]

Second minor

[ \begin{vmatrix} 1 & 1\ -1 & -\lambda \end{vmatrix} =1(-\lambda)-1(-1)=1-\lambda ]

Third minor

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

Substitute back: [ \det(A)=(1-\lambda)(\lambda-1)^2+2(1-\lambda)-2(1-\lambda) ] The last two terms cancel, so [ \det(A)=(1-\lambda)(\lambda-1)^2 ] Since $(1-\lambda)=-(\lambda-1)$, [ \det(A)=-(\lambda-1)^3 ]

3. Condition for non-trivial solution

[ \det(A)=0 \iff -(\lambda-1)^3=0 \iff \lambda=1 ] Thus the set of all such values of $\lambda$ is [ {1} ] which is a singleton.

4. Check options

• A: empty set — false
• B: contains more than two elements — false
• C: is a singleton — true
• D: contains exactly two elements — false

5. Final answer

The correct option is C.