"Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}, \vec{c}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{d}$ be a vector such that $\vec{b} \times \vec{d}=\vec{c} \times \vec{d}$ and $\vec{a} \cdot \vec{d}=4$. Then $|(\vec{a} \times \vec{d})|^2$ is equal to___________."

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Given Conditions :

The condition $\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$ implies that $(\vec{b} - \vec{c}) \times \vec{d} = 0$.

This indicates that $\vec{d}$ is parallel to $\vec{b} - \vec{c}$.

Expressing $\vec{b} - \vec{c}$:

Calculate $\vec{b} - \vec{c} = (3 \hat{i} - 3 \hat{j} + 3 \hat{k}) - (2 \hat{i} - \hat{j} + 2 \hat{k}) = \hat{i} - 2 \hat{j} + \hat{k}$.

Express $\vec{d}$:

Since $\vec{d}$ is parallel to $\vec{b} - \vec{c}$, let $\vec{d} = \lambda(\hat{i} - 2 \hat{j} + \hat{k})$.

Using the condition $\vec{a} \cdot \vec{d} = 4$:

Compute $\vec{a} \cdot \vec{d} = (\hat{i} + 2 \hat{j} + \hat{k}) \cdot (\lambda(\hat{i} - 2 \hat{j} + \hat{k}))$.

This simplifies to $\lambda(1 \cdot 1 + 2 \cdot (-2) + 1 \cdot 1) = \lambda(1 - 4 + 1) = -2\lambda$.

Set $-2\lambda = 4$, which gives $\lambda = -2$.

Determine $\vec{d}$ using $\lambda$:

Therefore, $\vec{d} = -2(\hat{i} - 2 \hat{j} + \hat{k}) = -2 \hat{i} + 4 \hat{j} - 2 \hat{k}$.

Calculate $\vec{a} \times \vec{d}$:

Use the determinant form for the cross product:

$ \vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 4 & -2 \end{vmatrix} $

Calculate the determinant:

For $\hat{i}$: $(2 \times -2) - (1 \times 4) = -4 - 4 = -8$

For $\hat{j}$: $(-(1 \times -2) - (1 \times -2)) = 2 - 2 = 0$

For $\hat{k}$: $(1 \times 4) - (2 \times -2) = 4 + 4 = 8$

Thus, $\vec{a} \times \vec{d} = -8 \hat{i} + 0 \hat{j} + 8 \hat{k}$.

Calculate the magnitude squared:

Find $|\vec{a} \times \vec{d}|^2 = (-8)^2 + 0^2 + 8^2 = 64 + 0 + 64 = 128$.

Therefore, $|(\vec{a} \times \vec{d})|^2 = 128$.

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Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}, \vec{c}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{d}$ be a vector such that $\vec{b} \times \vec{d}=\vec{c} \times \vec{d}$ and $\vec{a} \cdot \vec{d}=4$. Then $|(\vec{a} \times \vec{d})|^2$ is equal to___________.

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Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}, \vec{c}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{d}$ be a vector such that $\vec{b} \times \vec{d}=\vec{c} \times \vec{d}$ and $\vec{a} \cdot \vec{d}=4$. Then $|(\vec{a} \times \vec{d})|^2$ is equal to___________.

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