JEE Math Practice Question

Let $$\mathrm{d}$$ be the distance of the point of intersection of the lines $$\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$$ and $$\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$$ from the point $$(7,8,9)$$. Then $$\mathrm{d}^2+6$$ is equal to :

1. A.75
2. B.78
3. C.72
4. D.69