Vector Condition Analysis

Given vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=3\hat{i}+2\hat{j}-\hat{k}$, $\vec{c} = \lambda\hat{j} + \mu\hat{k}$, and a unit vector $\vec{d}$.

The condition $\vec{a}\times\vec{d}=\vec{b}\times\vec{d}$ implies $(\vec{a}-\vec{b})\times\vec{d} = \vec{0}$. This means $\vec{d}$ is parallel to $\vec{a}-\vec{b}$.

Calculating $\vec{a}-\vec{b}$: $ \vec{a}-\vec{b} = (\hat{i}+\hat{j}+\hat{k}) - (3\hat{i}+2\hat{j}-\hat{k}) = -2\hat{i} - \hat{j} + 2\hat{k} $

The magnitude is $|\vec{a}-\vec{b}| = \sqrt{(-2)^2 + (-1)^2 + (2)^2} = \sqrt{4+1+4} = 3$. Since $\vec{d}$ is a unit vector, $|\vec{d}|=1$. Therefore, $\vec{d} = \pm \frac{1}{3}(\vec{a}-\vec{b}) = \pm \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k})$.

The condition that $\vec{c}$ is perpendicular to $\vec{a}$ means their dot product is zero: $\vec{c}\cdot\vec{a} = 0$. $ (\lambda\hat{j} + \mu\hat{k}) \cdot (\hat{i}+\hat{j}+\hat{k}) = 0(\lambda) + 1(\lambda) + 1(\mu) = \lambda + \mu $. Setting $\lambda + \mu = 0$, we find $\mu = -\lambda$.

This simplifies the expression for $\vec{c}$ to $\vec{c} = \lambda\hat{j} - \lambda\hat{k} = \lambda(\hat{j}-\hat{k})$.

Scalar Value Determination

Use the condition $\vec{c}\cdot\vec{d} = 1$. We can use the expression for $\vec{d}$ derived earlier. Let's choose $\vec{d} = \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k})$. The final result will be independent of the sign choice.

$ \vec{c}\cdot\vec{d} = [\lambda(\hat{j}-\hat{k})] \cdot [\frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k})] = 1 $.

Performing the dot product: $ \frac{\lambda}{3} [ (0)(-2) + (1)(-1) + (-1)(2) ] = 1 $

$ \frac{\lambda}{3} [ 0 - 1 - 2 ] = 1 $

$ \frac{\lambda}{3} [-3] = 1 \implies -\lambda = 1 $. Thus, $\lambda = -1$.

Since $\mu = -\lambda$, we calculate $\mu = -(-1) = 1$.

Final Magnitude Calculation

The goal is to find the value of $|3 \lambda\vec{d} + \mu\vec{c}|^2$. Substitute the determined values $\lambda = -1$ and $\mu = 1$. The expression becomes $|3 (-1)\vec{d} + (1)\vec{c}|^2 = |-3\vec{d} + \vec{c}|^2$.

Calculate $-3\vec{d}$: $ -3\vec{d} = -3 \times \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k}) = -(-2\hat{i} - \hat{j} + 2\hat{k}) = 2\hat{i} + \hat{j} - 2\hat{k} $.

Calculate $\vec{c}$ using $\lambda=-1$: $ \vec{c} = \lambda(\hat{j}-\hat{k}) = -1(\hat{j}-\hat{k}) = -\hat{j}+\hat{k} $.

Now compute the sum $-3\vec{d} + \vec{c}$: $ (2\hat{i} + \hat{j} - 2\hat{k}) + (-\hat{j}+\hat{k}) = 2\hat{i} + (1-1)\hat{j} + (-2+1)\hat{k} = 2\hat{i} - \hat{k} $.

Finally, calculate the squared magnitude: $ |2\hat{i} - \hat{k}|^2 = (2)^2 + (0)^2 + (-1)^2 = 4 + 0 + 1 = 5 $.