Let's write the line in a slightly different vector form:
#(x,y,z) = (0, 6,1)+ t(3hati-4hatj-8hatk)#
Because the line is within the plane, we know that the vector, #veca= 3hati-4hatj-8hatk# is in the plane
Let #t=0# and we find that the point #(0,6,1) # is in the plane.
We can use the two points #(0,6,1) # and #(1,2,3)# to find another vector in the plane:
#vecb = (1-0)hati+(2-6)hatj+(3-1)hatk#
#vecb = hati-4hatj+2hatk#
We can use vectors #veca# and #vecb# to find a normal vector, #vecn#, to the plane:
#vecn = vecb xx veca#
I shall use a determinant with 5 columns to compute the cross product:
#vecn = |
(hati,hatj,hatk,hati,hatj),
(1,-4,2,1,-4),
(3,-4,-8,3,-4)
| = {(-4)(-8)-(2)(-4)}hati+{(2)(3)-(1)(-8)}hatj+{(1)(-4)-(-4)(3)}hatk#
#vecn = 40hati+14hatj+8hatk#
Because #vecn = 40hati+14hatj+8hatk# is a normal vector to the plane, we know that the equation of the plane is of the form:
#40x+14y+8z = C#
where C is a constant that we determine using one of the points.
Substitute the point #(1,2,3)# into the equation:
#40(1)+14(2)+8(3) = C#
#C = 92#
The equation of the plane is:
#40x+14y+8z = 92#
We can simplify this a bit by dividing both sides by 2:
#20x+7y+4z = 46#
You can check that line is in the plane by checking the point where t = 0, #(0,6,1)# and the point where #t = 1#, #(3,2,-7)#.
