
Yefim S. answered 04/20/21
Math Tutor with Experience
Let get 2 points on this line: x = z + 1 and y = 3 - 2z;
z = 0; x =1, y = 3; point B(1, 3, 0)
z = 1; x = 2, y = 1; point C(2,1,1)
And given point A(-1,2,1)
Vector AB = <1 + 1, 3 - 2, 0 - 1> = <2, 1, - 1>
Vector AC = <2 + 1, 1 - 2,1 - 1> = <3, - 1, 0>
Ii j k I
AB×AC = I2 1 -1I = - i - 3j - 5k is normal vector to these plane
I 3 -1 0I]
Now equation of plane: - 1(x - 1) - 3(y - 3) - 5(z - 0) = 0; x + 3y + 5z - 10 = 0