You need to know: The equation of a plane containing (x_0,y_0,z_0), perpendicular to the vector Ai+ Bj+ Ck, is A(x-x_0)+ B(y- y_0)+ C(z- z_0)= 0. (For any point (x,y,z) in that plane, (x-x_0)i+ (y- y_0)j+ (z- z_0)k is in the plane and so is perpendicular to Ai+ Bj+ Ck. The dot product of the two is 0.)
In this problem, the vector PQ is (3-1)i+ (1-(-2))j+ (2-(-4))k= 2i+ 3j+ 6k
As far as (b) is concerned, I'll bet there is a formula for the distance between a point and a plane in this same section of your textbook. To do it without that formula, remember that the shortest distance between a point and a plane is along a line perpendicular to that plane.
i) Write the parametric equations, for x, y, z in terms of a paratmeter t, for a line through (-1, 1, 1) in the same direction as the vector 2i+ 3j+ 6k.
ii) Plug those equations into the equation of the plane, from (a), to get a single equation in the single variable t and solve for t.
iii) Put that value of t into the parametric equations to get (x,y,z) coordinates of the point on the line and plane.
iii) Calculate the distance between that point and (-1, 1, 1).
