Let $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$, $$\vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k}$$ and $$\vec{c} = x\hat{i} + 2\hat{j} + 3\hat{k}$$, $$x \in \mathbb{R}$$. If $$\vec{d}$$ is the unit vector in the direction of $$\vec{b} + \vec{c}$$ such that $$\vec{a} \cdot \vec{d} = 1$$, then $$(\vec{a} \times \vec{b}) \cdot \vec{c}$$ is equal to

Given $$\vec{a}=(1,1,1)$$, $$\vec{b}=(2,4,-5)$$, and $$\vec{c}=(x,2,3)$$, we seek $$(\vec{a}\times\vec{b})\cdot\vec{c}$$ under the constraint that $$\vec{d}$$ is a unit vector in the direction of $$\vec{b}+\vec{c}$$ and satisfies $$\vec{a}\cdot\vec{d}=1$$.

Since $$\vec{d} = \frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|}$$ and $$\vec{b}+\vec{c} = (2+x,6,-2)$$, we have $$\vec{a}\cdot\vec{d} = \frac{(1)(2+x)+(1)(6)+(1)(-2)}{|\vec{b}+\vec{c}|} = \frac{x+6}{|\vec{b}+\vec{c}|} = 1$$.

Since $$|\vec{b}+\vec{c}| = \sqrt{(2+x)^2 + 36 + 4} = \sqrt{(2+x)^2 + 40}$$, substituting into the previous equation gives $$x+6 = \sqrt{(x+2)^2 + 40}$$. Squaring both sides yields $$(x+6)^2 = (x+2)^2 + 40$$, which simplifies to $$x^2 + 12x + 36 = x^2 + 4x + 44$$. This leads to $$8x = 8$$ and hence $$x = 1$$.

With $$x = 1$$, the vector $$\vec{c}$$ becomes $$(1,2,3)$$. Therefore the scalar triple product is $$[\vec{a},\vec{b},\vec{c}] = \begin{vmatrix}1 & 1 & 1\\ 2 & 4 & -5\\ 1 & 2 & 3\end{vmatrix}$$. Expanding gives $$1(12+10) - 1(6+5) + 1(4-4) = 22 - 11 + 0 = 11$$.

The correct answer is Option (1): 11.