If $$\lambda$$ be the ratio of the roots of the quadratic equation in x, $$3m^2x^2 + m(m-4)x + 2 = 0$$, then the least value of m for which $$\lambda + \frac{1}{\lambda} = 1$$, is:

We are given the quadratic equation in the variable $$x$$

$$3m^{2}x^{2}+m(m-4)x+2=0$$

Let the two (non-zero) roots of this equation be $$\alpha$$ and $$\beta$$. Their ratio is denoted by $$\lambda$$, so we write

$$\lambda=\dfrac{\alpha}{\beta}\qquad\Longrightarrow\qquad \lambda+\dfrac1{\lambda}=\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha} =\dfrac{\alpha^{2}+\beta^{2}}{\alpha\beta}.$$

The sum and product of the roots of any quadratic $$ax^{2}+bx+c=0$$ are given by the standard relations

$$\alpha+\beta=-\dfrac{b}{a},\qquad \alpha\beta=\dfrac{c}{a}.$$

For our quadratic we have the coefficients

$$a=3m^{2},\qquad b=m(m-4)=m^{2}-4m,\qquad c=2.$$

Therefore

$$\alpha+\beta=-\dfrac{m^{2}-4m}{3m^{2}} =\dfrac{4m-m^{2}}{3m^{2}},$$

$$\alpha\beta=\dfrac{2}{3m^{2}}.$$

To express $$\lambda+\dfrac1{\lambda}$$ in terms of the symmetric functions, first note the algebraic identity

$$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta.$$

Using this, we obtain

$$\lambda+\dfrac{1}{\lambda} =\dfrac{\alpha^{2}+\beta^{2}}{\alpha\beta} =\dfrac{(\alpha+\beta)^{2}-2\alpha\beta}{\alpha\beta}.$$

The condition given in the question is

$$\lambda+\dfrac1{\lambda}=1.$$

Substituting the above expression, we get

$$\dfrac{(\alpha+\beta)^{2}-2\alpha\beta}{\alpha\beta}=1.$$

Multiplying both sides by $$\alpha\beta$$ yields

$$(\alpha+\beta)^{2}-2\alpha\beta=\alpha\beta,$$ so that

$$(\alpha+\beta)^{2}=3\alpha\beta.$$

Now we substitute the explicit values of $$\alpha+\beta$$ and $$\alpha\beta$$ obtained from the coefficients.

First compute the square of the sum:

$$(\alpha+\beta)^{2} =\left(\dfrac{4m-m^{2}}{3m^{2}}\right)^{2} =\dfrac{(4m-m^{2})^{2}}{9m^{4}} =\dfrac{m^{2}(4-m)^{2}}{9m^{4}} =\dfrac{(4-m)^{2}}{9m^{2}}.$$

The product is

$$\alpha\beta=\dfrac{2}{3m^{2}}.$$

Setting $$(\alpha+\beta)^{2}=3\alpha\beta$$ gives

$$\dfrac{(4-m)^{2}}{9m^{2}}=3\left(\dfrac{2}{3m^{2}}\right).$$

Simplifying the right-hand side:

$$3\left(\dfrac{2}{3m^{2}}\right)=\dfrac{2}{m^{2}}.$$

Hence we have

$$\dfrac{(4-m)^{2}}{9m^{2}}=\dfrac{2}{m^{2}}.$$

Because $$m\neq0$$, the factors $$m^{2}$$ cancel out, leaving

$$(4-m)^{2}=18.$$

Taking square roots on both sides,

$$4-m=\pm\sqrt{18}=\pm3\sqrt{2}.$$

Thus the possible values of $$m$$ are

$$m=4-3\sqrt{2}\quad\text{or}\quad m=4+3\sqrt{2}.$$

Among these, the least value is clearly

$$m=4-3\sqrt{2}.$$

This value exactly matches option D in the list provided.

Hence, the correct answer is Option D.