On which of the following lines lies the point of intersection of the line, $$\frac{x-4}{2} = \frac{y-5}{2} = \frac{z-3}{1}$$ and the plane, $$x + y + z = 2$$?

We are given the straight line $$\dfrac{x-4}{2}=\dfrac{y-5}{2}=\dfrac{z-3}{1}$$ and the plane $$x+y+z=2$$. To find their point of intersection we first write the symmetric form of the line in parametric form.

Let the common ratio be the parameter $$\lambda$$. Then

$$\dfrac{x-4}{2}=\lambda,\qquad \dfrac{y-5}{2}=\lambda,\qquad \dfrac{z-3}{1}=\lambda.$$

So we obtain the parametric equations

$$x=4+2\lambda,\qquad y=5+2\lambda,\qquad z=3+\lambda.$$

Any point on the line therefore has coordinates $$\bigl(4+2\lambda,\;5+2\lambda,\;3+\lambda\bigr).$$

Now we impose the condition that this point also lies on the plane $$x+y+z=2$$. Substituting the above expressions of $$x,y,z$$ into the plane equation, we get

$$\bigl(4+2\lambda\bigr)+\bigl(5+2\lambda\bigr)+\bigl(3+\lambda\bigr)=2.$$

Simplifying step by step, we first add the constant terms:

$$4+5+3=12,$$

and then add the coefficients of $$\lambda$$:

$$(2\lambda+2\lambda+\lambda)=5\lambda.$$

Thus the substitution produces

$$12+5\lambda=2.$$

Moving the constant term to the right side, we have

$$5\lambda = 2-12=-10,$$

so

$$\lambda=\dfrac{-10}{5}=-2.$$

Putting $$\lambda=-2$$ back into the parametric equations of the line, we find the exact point of intersection:

$$x = 4+2(-2)=4-4=0,$$

$$y = 5+2(-2)=5-4=1,$$

$$z = 3+(-2)=1.$$

Hence the required point is $$P(0,\,1,\,1).$$

Our task is now to decide which option describes a line passing through this point. We check each option one by one by substituting $$x=0,\;y=1,\;z=1$$ into its symmetric form and seeing whether the three fractions become equal.

Option A: $$\dfrac{x-4}{1}=\dfrac{y-5}{1}=\dfrac{z-5}{-1}.$$ At $$P(0,1,1)$$ we obtain $$\dfrac{0-4}{1}=-4,\; \dfrac{1-5}{1}=-4,\; \dfrac{1-5}{-1}=4.$$ Because $$-4 \neq 4,$$ the fractions are not equal; $$P$$ is not on this line.

Option B: $$\dfrac{x-1}{1}=\dfrac{y-3}{2}=\dfrac{z+4}{-5}.$$ Substituting $$P(0,1,1)$$ gives $$\dfrac{0-1}{1}=-1,\; \dfrac{1-3}{2}=\dfrac{-2}{2}=-1,\; \dfrac{1+4}{-5}=\dfrac{5}{-5}=-1.$$ All three values are the same (-1), so the point indeed lies on this line.

Option C: $$\dfrac{x-2}{2}=\dfrac{y-3}{2}=\dfrac{z+3}{3}.$$ For $$P(0,1,1)$$ we find $$\dfrac{0-2}{2}=-1,\; \dfrac{1-3}{2}=-1,\; \dfrac{1+3}{3}=\dfrac{4}{3},$$ and $$-1\neq\dfrac{4}{3}.$$ So $$P$$ is not on this line.

Option D: $$\dfrac{x+3}{3}=\dfrac{4-y}{3}=\dfrac{z+1}{-2}.$$ At $$P(0,1,1)$$ we get $$\dfrac{0+3}{3}=1,\; \dfrac{4-1}{3}=1,\; \dfrac{1+1}{-2}=-1.$$ Because $$1\neq-1,$$ this line also does not contain $$P.$$

Only Option B satisfies the condition that all three ratios are equal, which confirms that the desired line is the one given in Option B.

Hence, the correct answer is Option B.