Find an equation of the plane. the plane through the point (−5, 9, 10) and perpendicular to the line x = 1 + t, y = 4t, z = 2 − 3t

The direction vector of the line
L: x=1+t, y=4t, z=2-3t
is <1,4,-3>
which is also the required normal vector of the plane.

Since the plane passes through point (-5,9,10), the required plane is :
&Pi; 1(x-(-5)+4(y-9)-3(z-10)=0
=>
&Pi; x+4y-3z=1