Final answer:
The equation of the plane parallel to x+4y-3z=1 and passing through (2,1,0) is x+4y-3z-6=0.
Step-by-step explanation:
To solve the mathematical problem completely, we need to find an equation of a plane that is parallel to the given plane x+4y-3z=1 and passes through the point (2,1,0). Since the plane is parallel to the given plane, it will have the same normal vector. From the given plane's equation, we can determine that its normal vector is (1, 4, -3).
In the Cartesian coordinate system, a plane equation can be written in the form Ax+By+Cz+D=0, where (A, B, C) is the normal vector. To find the value of D for our new plane, we substitute the coordinates of the given point (2,1,0) into the equation of the plane using its normal vector:
1(2) + 4(1) - 3(0) + D = 0
2 + 4 + D = 0
D = -6
Therefore, the equation of the plane is x + 4y - 3z - 6 = 0.