Given the vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 3 \hat{i} + 2 \hat{j} - \hat{k}$, along with $\vec{c} = \lambda \hat{j} + \mu \hat{k}$, and $\hat{d}$ being a unit vector such that $\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$ and $\vec{c} \cdot \hat{d} = 1$, we proceed as follows:

Determine $\hat{d}$:

Since $\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$, we have:

$ (\vec{a} - \vec{b}) \times \hat{d} = 0 $

Thus, $\hat{d}$ is parallel to $\vec{a} - \vec{b}$. Calculate:

$ \vec{a} - \vec{b} = (-2 \hat{i} + \hat{j} + 2 \hat{k}) $

Therefore, we can express $\hat{d}$ as:

$ \hat{d} = \frac{-2}{3} \hat{i} - \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k} $

Solve for $\lambda$ and $\mu$:

Using the condition $\vec{c} \cdot \hat{d} = 1$:

$ \frac{-\lambda}{3} + \frac{2 \mu}{3} = 1 $

Simplify to:

$ -\lambda + 2\mu = 3 \quad \text{...(i)} $

Since $\vec{c}$ is perpendicular to $\vec{a}$:

$ \vec{c} \cdot \vec{a} = 0 $

Which gives:

$ \lambda + \mu = 0 \quad \Rightarrow \lambda = -\mu $

Substitute $\lambda = -\mu$ into equation (i):

$ \mu + 2\mu = 3 \quad \Rightarrow 3\mu = 3 \quad \Rightarrow \mu = 1 $

Therefore, $\lambda = -1$.

Find $|3 \lambda \hat{d} + \mu \vec{c}|^2$:

Calculate:

$ |3 \lambda \hat{d} + \mu \vec{c}|^2 = 9 \lambda^2 |\hat{d}|^2 + \mu^2 |\vec{c}|^2 + 2 \cdot 3 \cdot \lambda \cdot \mu \cdot \vec{c} \cdot \vec{d} $

Substituting the determined values:

$ |\hat{d}|^2 = 1 $ (since $\hat{d}$ is a unit vector)

$ \vec{c} \cdot \vec{d} = 1 $

Thus:

$ = 9(-1)^2 \cdot 1 + 1^2 \cdot (1^2 + 1^2) + 2 \cdot 3 \cdot (-1) \cdot 1 \cdot 1 $

$ = 9 + 2 - 6 $

$ = 5 $

Hence, $|3 \lambda \hat{d} + \mu \vec{c}|^2 = 5$.