Let the given lines be $L_1$ and $L_2$. The equation of the lines are:

$L_1: \vec{r} = \hat{i} + \hat{j} + \hat{k} + p(2\hat{i} + \hat{j} + 2\hat{k})$

$L_2: \vec{r} = (1+q)\hat{i} + (1+q\lambda)\hat{j} + (1+q)\hat{k} = \hat{i} + \hat{j} + \hat{k} + q(\hat{i} + \lambda\hat{j} + \hat{k})$

The direction vectors of the lines $L_1$ and $L_2$ are $\vec{b_1} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b_2} = \hat{i} + \lambda\hat{j} + \hat{k}$ respectively.

Since the angle between the lines is $\frac{\pi}{2}$, the lines are perpendicular. Therefore, the dot product of their direction vectors is zero.

$\vec{b_1} \cdot \vec{b_2} = 0$

$(2\hat{i} + \hat{j} + 2\hat{k}) \cdot (\hat{i} + \lambda\hat{j} + \hat{k}) = 0$

$2(1) + 1(\lambda) + 2(1) = 0$

$2 + \lambda + 2 = 0$

$\lambda + 4 = 0$

$\lambda = -4$