Given:
P: (2,0,5)
L: (0,6,4)+t(7,-1,5)
and required plane, Π , passes through P and perpendicular to L.
The direction vector of L is V=<7,-1,5>.
For Π to be perpendicular to V, Π has V as the normal vector.
The equation of a plane with normal vector <7,-1,5> passing through a given point P(xp,yp,zp) is
7(x-xp)-1(y-yp)+5(z-zp)=0
Thus the equation of plane Π passing through P(2,0,5) is
7(x-2)-y+5(z-5)=0
or alternatively,
7x-y+5z = 14+25
7x-y+5z = 39
