​$$\text{For } \lambda \in \mathbb{R}, \text{ consider the system of differential equations:} \\x'_1 = x_1 + 2x_2 + 2x_3, \\x'_2 = 2x_2 + x_3, \\x'_3 = -x_3 + 2x_2 + \lambda x_3. \\[10pt]\text{If } \vec{x}(t) = \vec{a} \, t e^{2t} \, (\text{for some } \vec{a}) \text{ is a solution of the system, then the value of } \lambda \text{ is equal to.}$$​​

Correct option is A

​$$x_1' = x_1 + 2x_2 + 2x_3, \\[10pt]x_2' = 2x_2 + x_3, \\[10pt]x_3' = -x_3 + 2x_2 + \lambda x_3, \\[10pt]\text{Matrix of this system is given by:} \\[10pt]A = \begin{bmatrix}1 & 2 & 2 \\0 & 2 & 1 \\0 & 2 & \lambda - 1\end{bmatrix}. \\[10pt]\text{The solution } \vec{x}(t) = \vec{a} \, t e^{2t} \text{ is valid only if } 2 \text{ is an eigenvalue with multiplicity 2.} \\[10pt]\text{So, } \text{trace}(A) = 2 + 2 + \lambda_3.\ (\text{sum of eigen values}) \\[10pt]\text{Sum of diagonal entries} =1 + 2 + (\lambda - 1) = 4 + \lambda_3 \\[10pt]\text{This gives: } \lambda = 2 + \lambda_3. \cdots (1) \\[10pt]\text{Next, compute } \det(A): \\[10pt]\det(A) = 2 \cdot 2 \cdot \lambda_3 = 4 (\lambda_3). \\[10pt]\text{Expanding: }2. (\lambda - 1) - 2 = 4 \lambda_3. \\[10pt]2\lambda - 2 - 2 = 4 \lambda_3, \implies \lambda=2\lambda_3+2 \cdots (2) \ \\[10pt]\text{Using (1) and (2), we get:} \\[10pt]2 + \lambda_3 = 2 \lambda_3 + 2. \\[10pt] \implies \lambda_3=0\\[10pt]\text{Simplifying: } \lambda = 2. \\[10pt]\boxed{\lambda = 2}.$$​​