If the angle between the lines $$\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$$ and $$\frac{5-x}{-2} = \frac{7y-14}{P} = \frac{z-3}{4}$$ is $$\cos^{-1}\left(\frac{2}{3}\right)$$, then P is equal to:

We start with the first line, written as $$\dfrac{x}{2}=\dfrac{y}{2}=\dfrac{z}{1}\,.$$

When three fractions are equal to the same parameter (let us call it $$t$$), we can write

$$x = 2t,\qquad y = 2t,\qquad z = 1\cdot t.$$

Hence the direction ratios (d.r.’s) of the first line are plainly $$\bigl(2,\,2,\,1\bigr).$$

Now look at the second line: $$\dfrac{5-x}{-2}=\dfrac{7y-14}{P}=\dfrac{z-3}{4}\,.$$

Put all three fractions equal to the same parameter, say $$\lambda$$:

$$\dfrac{5-x}{-2}=\lambda,\qquad\dfrac{7y-14}{P}=\lambda,\qquad\dfrac{z-3}{4}=\lambda.$$

From the first fraction we get

$$5-x=-2\lambda\;\Longrightarrow\;x=5+2\lambda.$$

From the second fraction,

$$7y-14=P\lambda\;\Longrightarrow\;y=\dfrac{P\lambda+14}{7}=2+\dfrac{P}{7}\lambda.$$

From the third fraction,

$$z-3=4\lambda\;\Longrightarrow\;z=3+4\lambda.$$

The coefficients of $$\lambda$$ in these expressions give the direction ratios of the second line, namely

$$\bigl(2,\,\dfrac{P}{7},\,4\bigr).$$

To find the angle $$\theta$$ between two lines with direction ratios $$(a_1,b_1,c_1)$$ and $$(a_2,b_2,c_2)$$, we use the cosine formula

$$\cos\theta=\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2}}\;\sqrt{a_2^{2}+b_2^{2}+c_2^{2}}}\,.$$

For our lines we have

$$\begin{aligned} (a_1,b_1,c_1)&=(2,2,1),\\ (a_2,b_2,c_2)&=\left(2,\dfrac{P}{7},4\right). \end{aligned}$$

First compute the dot product (numerator):

$$a_1a_2+b_1b_2+c_1c_2 =2\cdot2+2\cdot\dfrac{P}{7}+1\cdot4 =4+\dfrac{2P}{7}+4 =8+\dfrac{2P}{7}.$$

Next, the magnitude of the first direction vector:

$$\sqrt{2^{2}+2^{2}+1^{2}} =\sqrt{4+4+1} =\sqrt{9}=3.$$

The magnitude of the second direction vector:

$$\sqrt{2^{2}+\left(\dfrac{P}{7}\right)^{2}+4^{2}} =\sqrt{4+\dfrac{P^{2}}{49}+16} =\sqrt{20+\dfrac{P^{2}}{49}} =\sqrt{\dfrac{980+P^{2}}{49}} =\dfrac{\sqrt{980+P^{2}}}{7}.$$

Therefore

$$\cos\theta=\dfrac{8+\dfrac{2P}{7}}{3\left(\dfrac{\sqrt{980+P^{2}}}{7}\right)} =\dfrac{8+\dfrac{2P}{7}}{3}\cdot\dfrac{7}{\sqrt{980+P^{2}}} =\dfrac{56+2P}{3\sqrt{980+P^{2}}}.$$

According to the question, the angle between the two lines satisfies $$\cos\theta=\dfrac{2}{3}.$$ Hence we must have

$$\dfrac{56+2P}{3\sqrt{980+P^{2}}}=\dfrac{2}{3}\,.$$

Multiplying both sides by $$3$$ gives

$$\dfrac{56+2P}{\sqrt{980+P^{2}}}=2.$$

Dividing by $$2$$ yields

$$\dfrac{56+2P}{2\sqrt{980+P^{2}}}=1 \;\Longrightarrow\; 56+2P=2\sqrt{980+P^{2}}.$$

Now divide by $$2$$ on both sides:

$$28+P=\sqrt{980+P^{2}}.$$

We square both sides to remove the square root:

$$(28+P)^{2}=980+P^{2}.$$

Expanding the left-hand side using $$(a+b)^{2}=a^{2}+2ab+b^{2}$$ gives

$$P^{2}+56P+784=980+P^{2}.$$

The $$P^{2}$$ terms cancel immediately, so

$$56P+784=980.$$

Subtract $$784$$ from both sides:

$$56P=196.$$

Finally divide by $$56$$:

$$P=\dfrac{196}{56}=3.5=\dfrac{7}{2}.$$

Among the given choices, $$\dfrac{7}{2}$$ corresponds to option B.

Hence, the correct answer is Option B.