$\overrightarrow{AB}=\left(4-2\right)\hat{i}+\left(5-3\right)\hat{j}+\left(7-4\right)\hat{k}=2\hat{i}+2\hat{j}+3\hat{k}$

Direction ratios 

$\overrightarrow{CD}=2\hat{i}+2\hat{j}+(k-3)\hat{k}$

Direction ratios

For $AB$ to be parallel to $CD$, the modulus of their direction cosines should be equal.

Direction cosines of $AB$ will be $\left(\frac{2}{\sqrt{17}},\frac{2}{\sqrt{17}},\frac{3}{\sqrt{17}}\right)$

And of $CD$ will be $\left(\frac{2}{\sqrt{8+{\left(k-3\right)}^2}},\frac{2}{\sqrt{8+{\left(k-3\right)}^2}},\frac{k-3}{\sqrt{8+{\left(k-3\right)}^2}}\right)$

So,

$\frac{2}{\sqrt{17}}=\left|\frac{2}{\sqrt{8+{\left(k-3\right)}^2}}\right|$

$\Rightarrow 17=8+{\left(k-3\right)}^2$

$\Rightarrow 9=k^2+9-6k$

$\Rightarrow k\left(k-6\right)=0$

$\Rightarrow k=6$