The equation of the plane containing two lines $\frac{x+1}{3}=\frac{y+3}{3}=\frac{z+5}{7}\text{\hspace{0.17em}and\hspace{0.17em}}\frac{x-1}{1}=\frac{y+3}{3}=\frac{z+7}{\lambda }is(\text{where\hspace{0.17em}}\lambda \in R-\left\{0\right\})$

$\frac{x+1}{3}=\frac{y+3}{3}=\frac{z+5}{7}and\frac{x-1}{1}=\frac{y+3}{3}=\frac{z+7}{\lambda }$

$(\overrightarrow{r}-(\widehat{i}-3\widehat{j}-7\widehat{k})).\overrightarrow{n}$

$\text{Where\hspace{0.17em}}\overrightarrow{a}=3\widehat{i}+3\widehat{j}+7\widehat{k};\overrightarrow{b}=2\widehat{i}-2\widehat{k}$

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 3& 3& 7\\ 2& 0& -2\end{array}\right|$

$\overrightarrow{\mathbf{n}}\mathbf{=}\mathit{\lambda }\mathbf{(}\mathbf{-}\mathbf{6}\widehat{\mathbf{i}}\mathbf{+}\mathbf{20}\widehat{\mathbf{j}}\mathbf{-}\mathbf{6}\widehat{\mathbf{k}}\mathbf{)}$

$(\overrightarrow{r}.(\widehat{i}-3\widehat{j}-7\widehat{k}))(3\widehat{i}-10\widehat{j}+3\widehat{k})$

$3x-10y+3z-12=0$