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 __________.

Given
$$\vec{a}= \hat{i}+ \hat{j}+ \hat{k},\; \vec{b}=3\hat{i}+2\hat{j}-\hat{k},\; \vec{c}= \lambda\hat{j}+ \mu\hat{k},\; \hat{d}\; \text{(unit)}$$
with the conditions
$$\vec{a}\times\hat{d}=\vec{b}\times\hat{d},\qquad \vec{c}\cdot\hat{d}=1,\qquad \vec{c}\perp\vec{a}.$$

Step 1: Relating $$\vec{a}$$ and $$\vec{b}$$ to $$\hat{d}$$
The identity $$\vec{u}\times\hat{d}=\vec{v}\times\hat{d}\;\Longrightarrow\;(\vec{u}-\vec{v})\times\hat{d}= \vec{0}$$ means $$\vec{u}-\vec{v}$$ is parallel to $$\hat{d}$$.
Thus $$\bigl(\vec{a}-\vec{b}\bigr)\times\hat{d}= \vec{0}\;\Longrightarrow\;\vec{a}-\vec{b}=t\hat{d}\;$$ for some scalar $$t$$.

Compute $$\vec{a}-\vec{b}$$:
$$\vec{a}-\vec{b}=(1-3)\hat{i}+(1-2)\hat{j}+(1-(-1))\hat{k}=-2\hat{i}-\hat{j}+2\hat{k}.$$

The magnitude of this vector is
$$\lvert\vec{a}-\vec{b}\rvert=\sqrt{(-2)^2+(-1)^2+2^2}=3.$$

Therefore the required unit vector is
$$\hat{d}=s\,\frac{-2\hat{i}-\hat{j}+2\hat{k}}{3},\qquad s=\pm1.$$
Explicitly
$$\hat{d}=s\Bigl(-\tfrac23\,\hat{i}-\tfrac13\,\hat{j}+\tfrac23\,\hat{k}\Bigr).$$

Step 2: Using $$\vec{c}\cdot\hat{d}=1$$
With $$\vec{c}=(0,\lambda,\mu)$$, dot product gives
$$\vec{c}\cdot\hat{d}=s\Bigl(-\tfrac{\lambda}{3}+\tfrac{2\mu}{3}\Bigr)=1$$
$$\Longrightarrow\;s(-\lambda+2\mu)=3\;.$$ $$-(1)$$

Step 3: Using $$\vec{c}\perp\vec{a}$$
Orthogonality with $$\vec{a}=(1,1,1)$$ yields
$$\vec{a}\cdot\vec{c}=0\;\Longrightarrow\;\lambda+\mu=0\;\Longrightarrow\;\mu=-\lambda.$$ $$-(2)$$

Step 4: Determining $$\lambda,\mu$$
Substituting $$\mu=-\lambda$$ from $$(2)$$ into $$(1)$$:
$$s(-\lambda+2(-\lambda))=s(-3\lambda)=3\;\Longrightarrow\;\lambda=-\frac{1}{s},\qquad \mu=-\lambda=\frac{1}{s}.$$

Hence
For $$s=1:\; \lambda=-1,\;\mu=1,\;\hat{d}=(-\tfrac23,-\tfrac13,\tfrac23).$$
For $$s=-1:\;\lambda=1,\;\mu=-1,\;\hat{d}=(\tfrac23,\tfrac13,-\tfrac23).$$

Step 5: Evaluating $$\bigl|\,3\lambda\hat{d}+\mu\vec{c}\bigr|^2$$
Take the case $$s=1$$ (the other choice will give the same result).
Here $$\lambda=-1,\;\mu=1$$ and
$$3\lambda\hat{d}=3(-1)\Bigl(-\tfrac23,-\tfrac13,\tfrac23\Bigr)=(2,1,-2),$$
$$\mu\vec{c}=1\,(0,-1,1)=(0,-1,1).$$

Therefore
$$3\lambda\hat{d}+\mu\vec{c}=(2,1,-2)+(0,-1,1)=(2,0,-1).$$

The square of its magnitude is
$$\lvert(2,0,-1)\rvert^{2}=2^{2}+0^{2}+(-1)^{2}=4+1=5.$$

Repeating with $$s=-1$$ also produces the vector $$(2,0,-1)$$, confirming the result is independent of the sign choice.

Final Answer:
$$\bigl|\,3\lambda\hat{d}+\mu\vec{c}\bigr|^{2}=5.$$