Let $$\hat{u}$$ and $$\hat{v}$$ be unit vectors inclined at acute angle such that $$|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}$$. If $$\vec{A} = \lambda\hat{u} + \hat{v} + (\hat{u} \times \hat{v})$$, then $$\lambda$$ is equal to :

Since $$\hat u,\;\hat v$$ are unit vectors, let the angle between them be $$\theta$$.
Given $$\left|\hat u \times \hat v\right| = \sin\theta = \frac{\sqrt3}{2} \; \Longrightarrow \; \theta = 60^{\circ}\;(\text{i.e. } \cos\theta = \tfrac12).$$

The vector in the question is
$$\vec A = \lambda\,\hat u + \hat v + (\hat u \times \hat v).$$

Take the dot product of $$\vec A$$ with $$\hat u$$:

$$\vec A \cdot \hat u = \lambda(\hat u \cdot \hat u) + (\hat v \cdot \hat u) + \bigl(\hat u \times \hat v\bigr)\cdot\hat u.$$

Here $$\hat u \cdot \hat u = 1,$$ $$\hat v \cdot \hat u = \cos\theta = \tfrac12,$$ and $$\bigl(\hat u \times \hat v\bigr)\cdot\hat u = 0$$ because the cross-product is perpendicular to both $$\hat u$$ and $$\hat v$$. Hence

$$\vec A \cdot \hat u = \lambda + \tfrac12 \quad -(1).$$

Next, dot $$\vec A$$ with $$\hat v$$:

$$\vec A \cdot \hat v = \lambda(\hat u \cdot \hat v) + (\hat v \cdot \hat v) + \bigl(\hat u \times \hat v\bigr)\cdot\hat v,$$

giving $$\vec A \cdot \hat v = \lambda\!\left(\tfrac12\right) + 1 + 0 = \tfrac12\lambda + 1 \quad -(2).$$

Denote $$\vec A \cdot \hat u = P$$ and $$\vec A \cdot \hat v = Q.$$
From $$(1):\; P = \lambda + \tfrac12 \;\Longrightarrow\; \lambda = P - \tfrac12 \quad -(3).$$

From $$(2):\; Q = \tfrac12\lambda + 1.$$ Substitute $$\lambda$$ from $$(3):$$

$$Q = \tfrac12(P - \tfrac12) + 1 = \tfrac12P - \tfrac14 + 1 = \tfrac12P + \tfrac34 \quad -(4).$$

We now express $$\lambda$$ as a linear combination of $$P$$ and $$Q$$.
Multiply $$(4)$$ by $$-\tfrac23$$ and add to $$\tfrac43 P$$:

$$\tfrac43 P - \tfrac23 Q = \tfrac43 P - \tfrac23\!\left(\tfrac12P + \tfrac34\right) = \tfrac43 P - \tfrac13 P - \tfrac12 = P - \tfrac12.$$

Using $$(3),$$ $$P - \tfrac12 = \lambda.$$ Therefore

$$\boxed{\lambda = \tfrac43(\vec A \cdot \hat u) - \tfrac23(\vec A \cdot \hat v)}.$$

This matches Option A.