The least value of $$(\cos^{2} \theta- 6\sin \theta \cos \theta + 3\sin^{2} \theta +2)$$ is

We need to find the minimum value of $$E = \cos^2\theta - 6\sin\theta\cos\theta + 3\sin^2\theta + 2$$.

Since $$\cos^2\theta = \frac{1+\cos 2\theta}{2}$$, $$\sin^2\theta = \frac{1-\cos 2\theta}{2}$$ and $$\sin\theta\cos\theta = \frac{\sin 2\theta}{2}$$, substituting these into the expression gives

$$E = \frac{1+\cos 2\theta}{2} - 6 \cdot \frac{\sin 2\theta}{2} + 3 \cdot \frac{1-\cos 2\theta}{2} + 2$$

which simplifies to

$$= \frac{1}{2} + \frac{\cos 2\theta}{2} - 3\sin 2\theta + \frac{3}{2} - \frac{3\cos 2\theta}{2} + 2$$

and further to

$$= 4 - \cos 2\theta - 3\sin 2\theta$$.

For an expression of the form $$a\cos\phi + b\sin\phi$$, the range is $$[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$$, so here with $$a = -1$$ and $$b = -3$$ we have

$$\sqrt{(-1)^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10}$$

and hence $$-\cos 2\theta - 3\sin 2\theta$$ varies between $$-\sqrt{10}$$ and $$\sqrt{10}$$.

Therefore the minimum value of $$E$$ is

$$E_{min} = 4 + (-\sqrt{10}) = 4 - \sqrt{10}$$

The correct answer is Option 1: $$4 - \sqrt{10}$$.