The equations of one of the straight lines which pass through the point $\left(1,3\right)$ and make an angle ${\tan }^{-1}\sqrt{2}$ with the straight line, $y+1=3\sqrt{2}x$ is:

The correct option is D

$4\sqrt{2}x+5y-\left(15+4\sqrt{2}\right)=0$

Explanation for correct option:

Let the equation of the straight line is $y=mx+c$

The straight line passes through $\left(1,3\right)$

$\therefore 3=m+c$

Given that the equations of one of the straight lines which pass through the point $\left(1,3\right)$ and make an angle ${\tan }^{-1}\sqrt{2}$ with the straight line, $y+1=3\sqrt{2}x$

From the given figure,

$$\begin{gathered} \therefore \tan \left({\tan }^{-1}\sqrt{2}\right)=\left|\frac{\left(m-3\sqrt{2}\right)}{\left(1+3m\sqrt{2}\right)}\right| \\ \Rightarrow \sqrt{2}=\left|\frac{\left(m-3\sqrt{2}\right)}{\left(1+3m\sqrt{2}\right)}\right| \end{gathered}$$

For positive

$$\begin{gathered} \Rightarrow \sqrt{2}=\frac{\left(m-3\sqrt{2}\right)}{\left(1+3m\sqrt{2}\right)} \\ \Rightarrow 6m+\sqrt{2}=m-3\sqrt{2} \\ \Rightarrow 5m=-4\sqrt{2} \\ \Rightarrow m=\frac{-4\sqrt{2}}{5} \end{gathered}$$

For negative

$$\begin{gathered} \Rightarrow 6m+\sqrt{2}=-m+3\sqrt{2} \\ \Rightarrow 7m=2\sqrt{2} \\ \Rightarrow m=\frac{2\sqrt{2}}{7} \end{gathered}$$

For $m=-\frac{4\sqrt{2}}{5}$

$$\begin{gathered} \therefore c=3-m \\ =\frac{15+4\sqrt{2}}{5} \end{gathered}$$

$$\begin{gathered} \therefore y=-\frac{4\sqrt{2}x}{5}+\frac{\left(15+4\sqrt{2}\right)}{5} \\ \Rightarrow 4\sqrt{2}x+5y-\left(15+4\sqrt{2}\right)=0 \end{gathered}$$

For $m=\frac{2\sqrt{2}}{7}$

$$\begin{gathered} \therefore c=3-m \\ =\frac{21-2\sqrt{2}}{5} \end{gathered}$$

$$\begin{gathered} \therefore y=\frac{2\sqrt{2}x}{7}+\frac{\left(21-2\sqrt{2}\right)}{7} \\ \Rightarrow 7y-2\sqrt{2}x-\left(21-2\sqrt{2}\right)=0 \end{gathered}$$

Hence, option $\left(\mathrm{D}\right)$ is correct.