Let $$L_1: \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(\hat{i} - \hat{j} + 2\hat{k})$$, $$\lambda \in R$$, $$L_2: \vec{r} = (\hat{j} - \hat{k}) + \mu(3\hat{i} + \hat{j} + p\hat{k})$$, $$\mu \in R$$ and $$L_3: \vec{r} = \delta(l\hat{i} + m\hat{j} + n\hat{k})$$, $$\delta \in R$$ be three lines such that $$L_1$$ is perpendicular to $$L_2$$ and $$L_3$$ is perpendicular to both $$L_1$$ and $$L_2$$. Then the point which lies on $$L_3$$ is

$$L_1$$ direction: $$(1, -1, 2)$$. $$L_2$$ direction: $$(3, 1, p)$$.

$$L_1 \perp L_2$$: $$3 - 1 + 2p = 0 \Rightarrow p = -1$$. $$L_2$$ direction: $$(3, 1, -1)$$.

$$L_3$$ direction $$(l, m, n)$$ perpendicular to both $$L_1$$ and $$L_2$$:

$$(l, m, n) = (1, -1, 2) \times (3, 1, -1) = (1-2, 6+1, 1+3) = (-1, 7, 4)$$.

$$L_3$$: $$\vec{r} = \delta(-\hat{i} + 7\hat{j} + 4\hat{k})$$. Points on $$L_3$$: $$(-\delta, 7\delta, 4\delta)$$.

For $$\delta = 1$$: $$(-1, 7, 4)$$.

The answer is Option (1): $$\boxed{(-1, 7, 4)}$$.