If the line $$\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-1}{-1}$$ intersects the plane $$2x + 3y - z + 13 = 0$$ at a point P and the plane $$3x + y + 4z = 16$$ at a point Q, then PQ is equal to

We have the line given in symmetric form as $$\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}$$. Let this common ratio be $$\lambda$$. Hence the parametric equations of the line are

$$x = 2 + 3\lambda,\qquad y = -1 + 2\lambda,\qquad z = 1 - \lambda.$$

First we find the point $$P$$ where the line meets the plane $$2x + 3y - z + 13 = 0$$. Substituting the parametric coordinates into the plane:

$$2x = 2(2 + 3\lambda) = 4 + 6\lambda,$$

$$3y = 3(-1 + 2\lambda) = -3 + 6\lambda,$$

$$-z = -(1 - \lambda) = -1 + \lambda.$$

Adding these along with the constant $$13$$ from the plane equation:

$$\bigl(4 + 6\lambda\bigr) + \bigl(-3 + 6\lambda\bigr) + \bigl(-1 + \lambda\bigr) + 13 = 0.$$

Combining the constant terms and the $$\lambda$$ terms separately, we get

$$\bigl(4 - 3 - 1 + 13\bigr) + \bigl(6\lambda + 6\lambda + \lambda\bigr) = 0$$

$$13 + 13\lambda = 0.$$

Thus $$\lambda = -1.$$ Substituting $$\lambda = -1$$ back into the parametric relations gives the coordinates of $$P$$:

$$x_P = 2 + 3(-1) = -1,$$

$$y_P = -1 + 2(-1) = -3,$$

$$z_P = 1 - (-1) = 2.$$

So $$P(-1,\,-3,\,2).$$

Next we locate the point $$Q$$ where the same line meets the plane $$3x + y + 4z = 16$$. Using the same parametric coordinates in this plane’s equation gives

$$3x = 3(2 + 3\lambda) = 6 + 9\lambda,$$

$$y = -1 + 2\lambda,$$

$$4z = 4(1 - \lambda) = 4 - 4\lambda.$$

Adding, we obtain

$$\bigl(6 + 9\lambda\bigr) + \bigl(-1 + 2\lambda\bigr) + \bigl(4 - 4\lambda\bigr) = 16.$$

Collecting constants and $$\lambda$$ terms,

$$(6 - 1 + 4) + (9\lambda + 2\lambda - 4\lambda) = 16,$$

$$9 + 7\lambda = 16,$$

$$7\lambda = 7,$$

$$\lambda = 1.$$

Substituting $$\lambda = 1$$ back, the coordinates of $$Q$$ are

$$x_Q = 2 + 3(1) = 5,$$

$$y_Q = -1 + 2(1) = 1,$$

$$z_Q = 1 - 1 = 0.$$

So $$Q(5,\,1,\,0).$$

We now compute the distance $$PQ$$ using the distance formula in three dimensions. For two points $$P(x_1,y_1,z_1)$$ and $$Q(x_2,y_2,z_2)$$, the distance is

$$PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}.$$

Here,

$$x_2 - x_1 = 5 - (-1) = 6,$$

$$y_2 - y_1 = 1 - (-3) = 4,$$

$$z_2 - z_1 = 0 - 2 = -2.$$

Hence,

$$PQ = \sqrt{6^2 + 4^2 + (-2)^2} = \sqrt{36 + 16 + 4} = \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}.$$

Hence, the correct answer is Option C.