We are given that $\epsilon$ and $\lambda$ are uncorrelated random variables with variances $Var[\epsilon] = 2$ and $Var[\lambda] = 3$. We need to find the variance of the linear combination $3\epsilon + \lambda$.

1. Problem Description

We are given that $\epsilon$ and $\lambda$ are uncorrelated random variables with variances $Var[\epsilon] = 2$ and $Var[\lambda] = 3$. We need to find the variance of the linear combination $3\epsilon + \lambda$.

2. Solution Steps

Since $\epsilon$ and $\lambda$ are uncorrelated, their covariance is zero, i.e., $Cov(\epsilon, \lambda) = 0$. We need to find $Var[3\epsilon + \lambda]$.

We know that for any constants $a$ and $b$ and random variables $X$ and $Y$:

$Var[aX + bY] = a^2Var[X] + b^2Var[Y] + 2abCov(X, Y)$

In our case, $X = \epsilon$, $Y = \lambda$, $a = 3$, and $b = 1$.

Thus, $Var[3\epsilon + \lambda] = 3^2Var[\epsilon] + 1^2Var[\lambda] + 2(3)(1)Cov(\epsilon, \lambda)$

Since $Cov(\epsilon, \lambda) = 0$, the last term vanishes.

$Var[3\epsilon + \lambda] = 9Var[\epsilon] + Var[\lambda]$

Substituting the given values, we get:

$Var[3\epsilon + \lambda] = 9(2) + 3 = 18 + 3 = 21$

3. Final Answer

$Var[3\epsilon + \lambda] = 21$