1. Find direction ratios of the two lines

The first line is $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ So its direction ratios are $\vec d_1=(2,2,1).$

The second line is $\frac{5-x}{-2}=\frac{7y-14}{p}=\frac{z-3}{4}.$ Rewrite each part in standard symmetric form.

Since $\frac{5-x}{-2}=\frac{x-5}{2},$ and $\frac{7y-14}{p}=\frac{7(y-2)}{p}=\frac{y-2}{p/7},$ we get $\frac{x-5}{2}=\frac{y-2}{p/7}=\frac{z-3}{4}.$ Hence direction ratios are $\vec d_2=\left(2,\frac p7,4\right).$

2. Use the formula for angle between two lines

If the angle between the lines is $\theta$, then $\cos\theta=\frac{|\vec d_1\cdot \vec d_2|}{|\vec d_1|\,|\vec d_2|}.$ Given $\theta=\cos^{-1}\left(\frac23\right),$ so $\cos\theta=\frac23.$

Now, $\vec d_1\cdot \vec d_2=2\cdot 2+2\cdot \frac p7+1\cdot 4=8+\frac{2p}{7}.$ Also, $|\vec d_1|=\sqrt{2^2+2^2+1^2}=3,$ $|\vec d_2|=\sqrt{2^2+\left(\frac p7\right)^2+4^2}=\sqrt{20+\frac{p^2}{49}}.$

Thus, $\frac{\left|8+\frac{2p}{7}\right|}{3\sqrt{20+\frac{p^2}{49}}}=\frac23.$ Cancel $3$: $\frac{\left|8+\frac{2p}{7}\right|}{\sqrt{20+\frac{p^2}{49}}}=2.$ Squaring both sides, $\left(8+\frac{2p}{7}\right)^2=4\left(20+\frac{p^2}{49}\right).$

3. Solve for $p$

Expand the left side: $64+\frac{32p}{7}+\frac{4p^2}{49}=80+\frac{4p^2}{49}.$ The $\frac{4p^2}{49}$ terms cancel: $64+\frac{32p}{7}=80.$ So, $\frac{32p}{7}=16$ $32p=112$ $p=\frac{112}{32}=\frac72.$

4. Check with options

$p=\frac72$ which matches Option A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer is also A.