We are given two vectors:

• $\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}$ where $\lambda > 0$.
• $\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}$.

We need to find the value of $\alpha + \beta + \gamma$ where the set of possible values for $\lambda$ is $(\alpha, \beta) - \{\gamma\}$.

Condition 1: Obtuse Angle with Vector $\vec{b}$

If vector $\vec{a}$ makes an obtuse angle with vector $\vec{b}$, their dot product must be negative ($\vec{a} \cdot \vec{b} < 0$).

Calculate the dot product:

$ \vec{a} \cdot \vec{b} = (\sqrt{2})(-\lambda^2) + (-1)(4\sqrt{2}) + (\lambda)(4\sqrt{2}) $ $ \vec{a} \cdot \vec{b} = -\sqrt{2}\lambda^2 - 4\sqrt{2} + 4\sqrt{2}\lambda $

Apply the condition $\vec{a} \cdot \vec{b} < 0$:

$ -\sqrt{2}\lambda^2 + 4\sqrt{2}\lambda - 4\sqrt{2} < 0 $

Divide by $-\sqrt{2}$ and reverse the inequality sign:

$ \lambda^2 - 4\lambda + 4 > 0 $

Factor the quadratic:

$ (\lambda - 2)^2 > 0 $

This inequality is true for all real values of $\lambda$ except $\lambda = 2$. Since we are given $\lambda > 0$, the possible values of $\lambda$ from this condition are $\lambda \in (0, \infty) - \{2\}$.

Condition 2: Angle with Positive z-axis

The angle $\theta$ is between $\vec{a}$ and the positive z-axis ($\hat{k}$). We are given $\frac{\pi}{6} < \theta < \frac{\pi}{2}$.

The cosine of the angle $\theta$ is given by:

$ \cos \theta = \frac{\vec{a} \cdot \hat{k}}{|\vec{a}| |\hat{k}|} $

Calculate the terms:

• $\vec{a} \cdot \hat{k} = (\sqrt{2})(0) + (-1)(0) + (\lambda)(1) = \lambda$.
• $|\vec{a}| = \sqrt{(\sqrt{2})^2 + (-1)^2 + \lambda^2} = \sqrt{2 + 1 + \lambda^2} = \sqrt{3 + \lambda^2}$.
• $|\hat{k}| = 1$.

Therefore,

$ \cos \theta = \frac{\lambda}{\sqrt{3 + \lambda^2}} $

Apply the angle condition $\frac{\pi}{6} < \theta < \frac{\pi}{2}$. This implies $0 < \cos \theta < \frac{\sqrt{3}}{2}$.

Substitute the expression for $\cos \theta$:

$ 0 < \frac{\lambda}{\sqrt{3 + \lambda^2}} < \frac{\sqrt{3}}{2} $

Since $\lambda > 0$, the left inequality $0 < \frac{\lambda}{\sqrt{3 + \lambda^2}}$ is always true.

Consider the right inequality:

$ \frac{\lambda}{\sqrt{3 + \lambda^2}} < \frac{\sqrt{3}}{2} $

Square both sides (valid as both sides are positive):

$ \frac{\lambda^2}{3 + \lambda^2} < \frac{3}{4} $

Cross-multiply:

$ 4\lambda^2 < 3(3 + \lambda^2) $ $ 4\lambda^2 < 9 + 3\lambda^2 $ $ \lambda^2 < 9 $

Since $\lambda > 0$, this implies $0 < \lambda < 3$. The possible values of $\lambda$ from this condition are $\lambda \in (0, 3)$.

Determining the Set of Possible $\lambda$ Values

We need to satisfy both conditions simultaneously:

• From Condition 1: $\lambda \in (0, \infty) - \{2\}$.
• From Condition 2: $\lambda \in (0, 3)$.

The intersection of these two sets is $\lambda \in (0, 3) - \{2\}$.

This matches the given form $(\alpha, \beta) - \{\gamma\}$.

• $\alpha = 0$
• $\beta = 3$
• $\gamma = 2$

Final Calculation

Calculate $\alpha + \beta + \gamma$:

$ \alpha + \beta + \gamma = 0 + 3 + 2 = 5 $