One way of going about this (not necessarily the best but one that

comes to mind) is the following.

All the points p(lambda)=(0,7/4,0)+lambda(4,-8,3) are on a line perpendicular

to the given plane. This is because the vector (4, -8, 3) is the gradient

of 4x-8y+3z+14, and hence is normal to the given plane.

Then a plane perpendicular to the given plane through the given point is:

Ax+By+Cz=K,

where:

1.) -A = K (force the given point (-1,0,0) onto the plane))

2.) (7/4) B = K (force the point p(0) onto the plane)

3.) 4A + (7/4-8)B +3C=K (force the point p(1) onto the plane)

Which defines all three coefficients in terms of K. So pick any

non-zero value for K, and the equation of the plane through

the three points follows.

RonL