If $$\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$$, then the value of $$\left|\hat{i} \times (\vec{a} \times \hat{i})\right|^2 + \left|\hat{j} \times (\vec{a} \times \hat{j})\right|^2 + \left|\hat{k} \times (\vec{a} \times \hat{k})\right|^2$$, is equal to:

We have the given vector $$\vec{a}=2\hat{i}+\hat{j}+2\hat{k}.$$

The required expression is

$$\left|\hat{i}\times(\vec{a}\times\hat{i})\right|^{2}\;+\;\left|\hat{j}\times(\vec{a}\times\hat{j})\right|^{2}\;+\;\left|\hat{k}\times(\vec{a}\times\hat{k})\right|^{2}.$$

To evaluate each term quickly, we shall use the well-known vector triple product formula:

$$\vec{p}\times(\vec{q}\times\vec{r})=\vec{q}\,(\vec{p}\cdot\vec{r})-\vec{r}\,(\vec{p}\cdot\vec{q}).$$

First we take $$\vec{p}=\hat{i},\;\vec{q}=\vec{a},\;\vec{r}=\hat{i}.$$ Applying the formula, we get

$$\hat{i}\times(\vec{a}\times\hat{i})=\vec{a}\,(\hat{i}\cdot\hat{i})-\hat{i}\,(\hat{i}\cdot\vec{a}).$$

Now, $$\hat{i}\cdot\hat{i}=1$$ and $$\hat{i}\cdot\vec{a}=2.$$ Substituting these,

$$\hat{i}\times(\vec{a}\times\hat{i})=\vec{a}-2\hat{i}.$$

Explicitly,

$$\vec{a}-2\hat{i}=(2\hat{i}+\hat{j}+2\hat{k})-2\hat{i}=0\hat{i}+1\hat{j}+2\hat{k}.$$

The magnitude squared of this vector is

$$|0\hat{i}+1\hat{j}+2\hat{k}|^{2}=0^{2}+1^{2}+2^{2}=0+1+4=5.$$

Next, we take $$\vec{p}=\hat{j},\;\vec{q}=\vec{a},\;\vec{r}=\hat{j}.$$ Using the same formula,

$$\hat{j}\times(\vec{a}\times\hat{j})=\vec{a}\,(\hat{j}\cdot\hat{j})-\hat{j}\,(\hat{j}\cdot\vec{a}).$$

Here, $$\hat{j}\cdot\hat{j}=1$$ and $$\hat{j}\cdot\vec{a}=1.$$ So,

$$\hat{j}\times(\vec{a}\times\hat{j})=\vec{a}-\hat{j}.$$ That is,

$$\vec{a}-\hat{j}=(2\hat{i}+\hat{j}+2\hat{k})-\hat{j}=2\hat{i}+0\hat{j}+2\hat{k}.$$

The corresponding magnitude squared is

$$|2\hat{i}+0\hat{j}+2\hat{k}|^{2}=2^{2}+0^{2}+2^{2}=4+0+4=8.$$

Finally, we choose $$\vec{p}=\hat{k},\;\vec{q}=\vec{a},\;\vec{r}=\hat{k}.$$ Again,

$$\hat{k}\times(\vec{a}\times\hat{k})=\vec{a}\,(\hat{k}\cdot\hat{k})-\hat{k}\,(\hat{k}\cdot\vec{a}).$$

Because $$\hat{k}\cdot\hat{k}=1$$ and $$\hat{k}\cdot\vec{a}=2,$$ we have

$$\hat{k}\times(\vec{a}\times\hat{k})=\vec{a}-2\hat{k}.$$

So explicitly,

$$\vec{a}-2\hat{k}=(2\hat{i}+\hat{j}+2\hat{k})-2\hat{k}=2\hat{i}+1\hat{j}+0\hat{k}.$$

The magnitude squared of this vector equals

$$|2\hat{i}+1\hat{j}+0\hat{k}|^{2}=2^{2}+1^{2}+0^{2}=4+1+0=5.$$

Adding all three obtained values, we get

$$5+8+5=18.$$

So, the answer is $$18$$.