1. Given vectors

$\vec u = 3\hat i - \hat j = (3,-1,0), \qquad \vec v = 2\hat i + \hat j - \lambda \hat k = (2,1,-\lambda), \quad \lambda>0$

The angle between them is

$\theta = \cos^{-1}\left(\frac{\sqrt5}{2\sqrt7}\right)$

So,

$\cos\theta = \frac{\sqrt5}{2\sqrt7}$

2. Use the dot product formula

$\cos\theta = \frac{\vec u\cdot \vec v}{|\vec u||\vec v|}$

First compute:

$\vec u\cdot \vec v = 3\cdot 2 + (-1)\cdot 1 + 0\cdot (-\lambda) = 6-1=5$

Also,

$|\vec u| = \sqrt{3^2+(-1)^2} = \sqrt{10}$

$|\vec v| = \sqrt{2^2+1^2+\lambda^2} = \sqrt{5+\lambda^2}$

Thus,

$\frac{5}{\sqrt{10}\sqrt{5+\lambda^2}} = \frac{\sqrt5}{2\sqrt7}$

3. Solve for $\lambda$

Cross-multiplying,

$5\cdot 2\sqrt7 = \sqrt5\,\sqrt{10}\,\sqrt{5+\lambda^2}$

Since

$\sqrt5\,\sqrt{10} = \sqrt{50} = 5\sqrt2,$

we get

$10\sqrt7 = 5\sqrt2\,\sqrt{5+\lambda^2}$

Divide by $5$:

$2\sqrt7 = \sqrt2\,\sqrt{5+\lambda^2}$

Square both sides:

$4\cdot 7 = 2(5+\lambda^2)$

$28 = 10 + 2\lambda^2$

$2\lambda^2 = 18$

$\lambda^2 = 9$

Since $\lambda>0$,

$\lambda = 3$

Hence,

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

4. Use orthogonal decomposition

We are given

$\vec v = \vec v_1 + \vec v_2$

where $\vec v_1 \parallel \vec u$ and $\vec v_2 \perp \vec u$.

This is the standard decomposition of $\vec v$ into components parallel and perpendicular to $\vec u$.

Since $\vec v_1 \perp \vec v_2$, by Pythagoras,

$|\vec v|^2 = |\vec v_1|^2 + |\vec v_2|^2$

So we only need $|\vec v|^2$.

5. Compute $|\vec v|^2$

$|\vec v|^2 = 2^2 + 1^2 + (-3)^2 = 4+1+9=14$

Therefore,

$|\vec v_1|^2 + |\vec v_2|^2 = 14$

6. Check options

• A: $\frac{23}{2}$
• B: $\frac{25}{2}$
• C: $10$
• D: $14$

So the correct option is

$\boxed{\text{D}}$