Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\lambda \hat{j}+\mu \hat{k}$ and $\hat{d}$ be a unit vector such that $\vec{a} \times \hat{d}=\vec{b} \times \hat{d}$ and $\vec{c} \cdot \hat{d}=1$. If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3 \lambda \hat{d}+\mu \vec{c}|^2$ is equal to________

<p>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:</p> <p><p><strong>Determine $\hat{d}$:</strong></p> <p>Since $\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$, we have:</p> <p>$ (\vec{a} - \vec{b}) \times \hat{d} = 0 $</p> <p>Thus, $\hat{d}$ is parallel to $\vec{a} - \vec{b}$. Calculate:</p> <p>$ \vec{a} - \vec{b} = (-2 \hat{i} + \hat{j} + 2 \hat{k}) $</p> <p>Therefore, we can express $\hat{d}$ as:</p> <p>$ \hat{d} = \frac{-2}{3} \hat{i} - \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k} $</p></p> <p><p><strong>Solve for $\lambda$ and $\mu$:</strong></p> <p>Using the condition $\vec{c} \cdot \hat{d} = 1$:</p> <p>$ \frac{-\lambda}{3} + \frac{2 \mu}{3} = 1 $</p> <p>Simplify to:</p> <p>$ -\lambda + 2\mu = 3 \quad \text{...(i)} $</p> <p>Since $\vec{c}$ is perpendicular to $\vec{a}$:</p> <p>$ \vec{c} \cdot \vec{a} = 0 $</p> <p>Which gives:</p> <p>$ \lambda + \mu = 0 \quad \Rightarrow \lambda = -\mu $</p> <p>Substitute $\lambda = -\mu$ into equation (i):</p> <p>$ \mu + 2\mu = 3 \quad \Rightarrow 3\mu = 3 \quad \Rightarrow \mu = 1 $</p> <p>Therefore, $\lambda = -1$.</p></p> <p><p><strong>Find $|3 \lambda \hat{d} + \mu \vec{c}|^2$:</strong></p> <p>Calculate:</p> <p>$ |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} $</p> <p>Substituting the determined values:</p></p> <p><p>$ |\hat{d}|^2 = 1 $ (since $\hat{d}$ is a unit vector)</p></p> <p><p>$ \vec{c} \cdot \vec{d} = 1 $</p> <p>Thus:</p> <p>$ = 9(-1)^2 \cdot 1 + 1^2 \cdot (1^2 + 1^2) + 2 \cdot 3 \cdot (-1) \cdot 1 \cdot 1 $</p> <p>$ = 9 + 2 - 6 $</p> <p>$ = 5 $</p></p> <p>Hence, $|3 \lambda \hat{d} + \mu \vec{c}|^2 = 5$.</p>