Let $$\vec{a} = 2\hat{i} - \hat{j} + 5\hat{k}$$ and $$\vec{b} = \alpha \hat{i} + \beta \hat{j} + 2\hat{k}$$. If $$\left(\left(\vec{a} \times \vec{b}\right) \times \hat{i}\right) \cdot \hat{k} = \frac{23}{2}$$, then $$\left|\vec{b} \times 2\hat{j}\right|$$ is equal to

Given,

$$\vec a=2\hat i-\hat j+5\hat k$$ and $$\vec b=\alpha\hat i+\beta\hat j+2\hat k$$

Now,

$$\vec a\times\vec b= \begin{vmatrix} \hat i&\hat j&\hat k\\ 2&-1&5\\ \alpha&\beta&2 \end{vmatrix}$$

$$=\hat i(-2-5\beta)-\hat j(4-5\alpha)+\hat k(2\beta+\alpha)$$

$$=(-2-5\beta)\hat i+(5\alpha-4)\hat j+(2\beta+\alpha)\hat k$$

Now compute

$$((\vec a\times\vec b)\times\hat i)\cdot\hat k$$

Using $$\hat i=(1,0,0),$$

if

$$\vec p=p_1\hat i+p_2\hat j+p_3\hat k,$$

then

$$\vec p\times\hat i=p_3\hat j-p_2\hat k$$

Hence,

$$((\vec a\times\vec b)\times\hat i)\cdot\hat k =-(5\alpha-4)$$

Given,

$$-(5\alpha-4)=\frac{23}{2}$$

$$-5\alpha+4=\frac{23}{2}$$

$$-10\alpha+8=23$$

$$-10\alpha=15$$

$$\alpha=-\frac32$$

Now,

$$\vec b\times2\hat j= \begin{vmatrix} \hat i&\hat j&\hat k\\ \alpha&\beta&2\\ 0&2&0 \end{vmatrix}$$

$$=-4\hat i+0\hat j+2\alpha\hat k$$

Substituting

$$\alpha=-\frac32,$$

$$\vec b\times2\hat j=-4\hat i-3\hat k$$

Therefore,

$$|\vec b\times2\hat j| =\sqrt{(-4)^2+(-3)^2}$$

$$=\sqrt{16+9}$$

$$=5$$

Hence, the required value is

$$\boxed{5}$$.