Final answer:
The equation of plane P that is parallel to plane Q and passes through the point (2,3,8) is 4x + 5y + 7z = 68. This is because parallel planes share the same normal vector, which in this case is given by the coefficients of the x, y, and z terms in the equation of plane Q.
Step-by-step explanation:
To find the equation of plane P that passes through the point (2,3,8) and is parallel to plane Q with the equation 4x+5y+7z=15, we first need to understand that parallel planes have the same normal vector. The coefficients of x, y, and z in the equation of plane Q represent the normal vector to plane Q, which is (4,5,7). Since plane P is parallel to plane Q, it will also have a normal vector of (4,5,7).
The general equation of a plane is given by Nx(x-x0) + Ny(y-y0) + Nz(z-z0) = 0, where (Nx, Ny, Nz) is the normal vector and (x0, y0, z0) is a point on the plane. In this case, substituting the normal vector (4,5,7) and the point (2,3,8), we get 4(x-2) + 5(y-3) + 7(z-8) = 0.
Expanding and simplifying, the equation of plane P can be expressed as 4x + 5y + 7z = 68.