Find the line which passing thorugh point $(9,-13,0)$ and is parallel to the intersection line of planes $-5x-5y-z=0,-5x+5y+z=3$.
My attempt :
Find the intersection line :
det$\begin{pmatrix} i&j&k\\ -5&-5&-1\\ -5&5&1\\ \end{pmatrix}=10\hat{j}-50\hat{k}.$
Denote $z=0 \implies -5x-5y=0,-5x+5y=3 \implies x=-\frac{3}{10},y=\frac{3}{10}$
The intersection line is : $\vec{r(t)}=(-\frac{3}{10},\frac{3}{10},0)+t(0,10,-50)$
Is the requested line is $\vec{r(t)}=(9,-13,0)+t(0,10,-50)$ ?