If the line, $$\frac{x - 3}{1} = \frac{y + 2}{-1} = \frac{z + \lambda}{-2}$$ lies in the plane, $$2x - 4y + 3z = 2$$, then the shortest distance between this line and the line, $$\frac{x - 1}{12} = \frac{y}{9} = \frac{z}{4}$$ is

Let the given line be represented in the symmetric form

$$\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z+\lambda}{-2}=t$$

so that every point on it is

$$x=3+t,\;y=-2-t,\;z=-\lambda-2t\;.-(1)$$

This line is known to lie entirely in the plane

$$2x-4y+3z=2\;.-(2)$$

Substituting the coordinates from $$-(1)$$ into the plane equation $$-(2)$$ gives

$$2(3+t)-4(-2-t)+3(-\lambda-2t)=2.$$

Simplifying term by term, first expand each bracket:

$$6+2t+8+4t-3\lambda-6t=2.$$

Combine like terms for $$t$$:

$$6+8+(2t+4t-6t)-3\lambda=2,$$

$$14+0t-3\lambda=2,$$

$$14-3\lambda=2.$$

Rearrange to isolate $$\lambda$$:

$$-3\lambda=2-14=-12,$$

$$\lambda=4.$$

Thus the first line is finally

$$\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z+4}{-2}.$$

Its direction vector is therefore

$$\mathbf{d}_1=\langle1,-1,-2\rangle.$$(3)

For the second line we write

$$\frac{x-1}{12}=\frac{y}{9}=\frac{z}{4}=s,$$

yielding the parametric coordinates

$$x=1+12s,\;y=9s,\;z=4s\;.-(4)$$

The direction vector of this line is

$$\mathbf{d}_2=\langle12,9,4\rangle.$$(5)

To compute the shortest distance between two possibly skew lines, we employ

$$\text{Distance}=\frac{\lvert(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2)\rvert}{\lvert\mathbf{d}_1\times\mathbf{d}_2\rvert},$$

where $$\mathbf{a}_1$$ and $$\mathbf{a}_2$$ are any chosen points on the first and second lines respectively.

Choose the convenient points obtained at $$t=0$$ in $$-(1)$$ and $$s=0$$ in $$-(4):$$

$$\mathbf{a}_1=(3,-2,-4),\quad\mathbf{a}_2=(1,0,0).$$(6)

The joining vector is

$$\mathbf{a}_2-\mathbf{a}_1=\langle1-3,\;0-(-2),\;0-(-4)\rangle=\langle-2,2,4\rangle.$$(7)

Compute the cross product of the direction vectors.

$$\mathbf{d}_1\times\mathbf{d}_2= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\[4pt] 1&-1&-2\\[4pt] 12&9&4 \end{vmatrix}.$$

Expanding the determinant:

$$\mathbf{d}_1\times\mathbf{d}_2= \mathbf{i}\bigl((-1)\cdot4-(-2)\cdot9\bigr) -\mathbf{j}\bigl(1\cdot4-(-2)\cdot12\bigr) +\mathbf{k}\bigl(1\cdot9-(-1)\cdot12\bigr).$$

Compute each component:

$$\mathbf{i}( -4+18)=14\mathbf{i},$$

$$-\mathbf{j}( 4+24)=-28\mathbf{j},$$

$$\mathbf{k}(9+12)=21\mathbf{k}.$$

Thus

$$\mathbf{d}_1\times\mathbf{d}_2=\langle14,-28,21\rangle.$$(8)

The magnitude of this vector is

$$\lvert\mathbf{d}_1\times\mathbf{d}_2\rvert =\sqrt{14^{2}+(-28)^{2}+21^{2}} =\sqrt{196+784+441} =\sqrt{1421}.$$(9)

Now evaluate the scalar triple product in the numerator:

$$(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2) =\langle-2,2,4\rangle\cdot\langle14,-28,21\rangle.$$

Take the dot product term by term:

$$(-2)(14)+(2)(-28)+(4)(21)=-28-56+84=0.$$(10)

Because the numerator equals zero, the entire fraction vanishes:

$$\text{Distance}= \frac{\lvert0\rvert}{\sqrt{1421}}=0.$$(11)

Therefore the two lines intersect, and the least separation is zero.

Hence, the correct answer is Option D.