I must write an equation for plane which goes through point $B(2; -3; 1)$ perpendicular to line $ \left\{ \begin{array}{c} x=2t-1 \\ y=7-3t \\ z=0 \end{array} \right. $
From that I obtain line direction vector (normal vector of line?) $v=(2; -3; 0)$ And my plane equation is $$2(x-2)-3(y+3)+0=0$$ obtained using formula $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$ (B values are with indexes)
I am pretty sure I have misunderstood something though.