1. Given vectors

$\vec a = 3\hat i - \hat j + 2\hat k$

So in component form,

$\vec a = (3,-1,2)$

We also have

$\vec b = \vec a \times (\hat i - 2\hat k)$

and

$\hat i - 2\hat k = (1,0,-2).$

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2. Compute $\vec b = \vec a \times (1,0,-2)$

Using determinant:

$$\vec b= \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & -1 & 2\\ 1 & 0 & -2 \end{vmatrix}$$ $$\vec b = \hat i\big((-1)(-2)-2\cdot 0\big)-\hat j\big(3(-2)-2\cdot 1\big)+\hat k\big(3\cdot 0-(-1)\cdot 1\big)$$ $$\vec b = \hat i(2)-\hat j(-6-2)+\hat k(1)$$ $$\vec b = 2\hat i+8\hat j+\hat k$$

So,

$\vec b=(2,8,1).$

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3. Compute $\vec c = \vec b \times \hat k$

Since $\hat k=(0,0,1)$,

$$\vec c= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 8 & 1\\ 0 & 0 & 1 \end{vmatrix}$$ $$\vec c = \hat i(8\cdot 1-1\cdot 0)-\hat j(2\cdot 1-1\cdot 0)+\hat k(2\cdot 0-8\cdot 0)$$ $$\vec c = 8\hat i-2\hat j$$

Thus,

$\vec c=(8,-2,0).$

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4. Find $\vec c-2\hat j$

$$\vec c-2\hat j = 8\hat i-2\hat j-2\hat j = 8\hat i-4\hat j$$

So,

$$\vec c-2\hat j=(8,-4,0).$$

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5. Projection of $\vec c-2\hat j$ on $\vec a$

The scalar projection of vector $\vec u$ on $\vec a$ is

$$\text{proj}_{\vec a}(\vec u)=\frac{\vec u\cdot \vec a}{|\vec a|}.$$

Here,

$$\vec u = (8,-4,0), \qquad \vec a=(3,-1,2).$$

First compute the dot product:

$$\vec u\cdot \vec a = 8\cdot 3 + (-4)(-1) + 0\cdot 2 = 24+4=28.$$

Now compute $|\vec a|$:

$$|\vec a|=\sqrt{3^2+(-1)^2+2^2}=\sqrt{9+1+4}=\sqrt{14}.$$

Therefore,

$$\text{projection} = \frac{28}{\sqrt{14}}=\frac{28\sqrt{14}}{14}=2\sqrt{14}.$$

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6. Check options

• A: $2\sqrt7$
• B: $3\sqrt7$
• C: $\sqrt{14}$
• D: $2\sqrt{14}$

Hence the correct option is

$\boxed{D:~2\sqrt{14}}$