Question

Find an equation of the plane. the plane through the point (−2, 6, 10) and perpendicular to the line x = 3 + t, y = 4t, z = 1 − 3t

Answer 1

Any plane perpendicular to a line has a normal vector equal to the direction vector of the line.

Thus the normal vector of the plane is the director vector of the line
L: (3,0,1)+t(1,4,-3)
or <1,4,-3>

Given that the plane passes through (-2,6,10), we conclude that the plane has the following equation:
1(x-(-2))+4(y-6)-3(z-10)=0
=>
&Pi; : x+4y-3z+8=0