Consider two vectors $$\vec{u} = 3\hat{i} - \hat{j}$$ and $$\vec{v} = 2\hat{i} + \hat{j} - \lambda\hat{k}$$, $$\lambda \gt 0$$. The angle between them is given by $$\cos^{-1}\left(\dfrac{\sqrt{5}}{2\sqrt{7}}\right)$$. Let $$\vec{v} = \vec{v}_1 + \vec{v}_2$$, where $$\vec{v}_1$$ is parallel to $$\vec{u}$$ and $$\vec{v}_2$$ is perpendicular to $$\vec{u}$$. Then the value $$|\vec{v}_1|^2 + |\vec{v}_2|^2$$ is equal to

The given vectors are $$\vec u = 3\hat i - \hat j$$ and $$\vec v = 2\hat i + \hat j - \lambda \hat k$$ with $$\lambda \gt 0$$.
The angle $$\theta$$ between them satisfies

$$\cos\theta = \dfrac{\vec u \cdot \vec v}{\lvert\vec u\rvert\,\lvert\vec v\rvert} = \dfrac{\sqrt5}{2\sqrt7}\,\,\,\,-(1)$$

First compute the numerator:
$$\vec u \cdot \vec v = (3)(2) + (-1)(1) + (0)(-\lambda) = 6 - 1 = 5$$

Next compute the magnitudes:
$$\lvert\vec u\rvert = \sqrt{3^{2} + (-1)^{2}} = \sqrt{9 + 1} = \sqrt{10}$$ $$\lvert\vec v\rvert = \sqrt{2^{2} + 1^{2} + \lambda^{2}} = \sqrt{5 + \lambda^{2}}$$

Insert these results in $$(1)$$:

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

Cross-multiply and simplify:
$$10\sqrt7 = \sqrt5\,\sqrt{10}\,\sqrt{5+\lambda^{2}}$$ $$10\sqrt7 = (\sqrt{5\cdot10})\,\sqrt{5+\lambda^{2}} = 5\sqrt2\,\sqrt{5+\lambda^{2}}$$ Divide by $$5$$:
$$2\sqrt7 = \sqrt2\,\sqrt{5+\lambda^{2}}$$ Square both sides:
$$(2\sqrt7)^{2} = (\sqrt2\,\sqrt{5+\lambda^{2}})^{2}$$ $$4\cdot7 = 2\,(5+\lambda^{2})$$ $$28 = 10 + 2\lambda^{2}$$ $$18 = 2\lambda^{2}$$ $$\lambda^{2} = 9 \Longrightarrow \lambda = 3 \quad(\text{since } \lambda \gt 0)$$

Thus $$\vec v = 2\hat i + \hat j - 3\hat k$$ and its magnitude is

$$\lvert\vec v\rvert^{2} = 2^{2} + 1^{2} + (-3)^{2} = 4 + 1 + 9 = 14\,\,\,\,-(2)$$

Now write $$\vec v = \vec v_1 + \vec v_2$$ where $$\vec v_1$$ is parallel to $$\vec u$$ and $$\vec v_2$$ is perpendicular to $$\vec u$$.
Because a parallel component and a perpendicular component are orthogonal, we have the Pythagorean relation

$$\lvert\vec v\rvert^{2} = \lvert\vec v_1\rvert^{2} + \lvert\vec v_2\rvert^{2}\,\,\,\,-(3)$$

Substituting the value from $$(2)$$ into $$(3)$$ gives

$$\lvert\vec v_1\rvert^{2} + \lvert\vec v_2\rvert^{2} = 14$$

Hence the required value is $$14$$.
Option B is correct.