= a^3 +3a^2-7a+4a^2+12a-28
=a^3 +7a^2+5a-28
Answer:
The answer is 9 1/15 aka answer A.
Step-by-step explanation:
Too lazy to explain. :p
Answer:
Step-by-step explanation:
The line can be written as ![\frac{x-1}{2}= \frac{y-3}{-1}= \frac{z-0}{1} =t](https://tex.z-dn.net/?f=%5Cfrac%7Bx-1%7D%7B2%7D%3D%20%5Cfrac%7By-3%7D%7B-1%7D%3D%20%5Cfrac%7Bz-0%7D%7B1%7D%20%3Dt)
Since the line is contained in the plane, if we take any two points of this line they will lie in the plane.
Give values for t =0 and 1
We get two points (x,y,z) as (1,3,0) and (3,2,1)
Now we have another non collinearpoint (1,1,1) (given)
The plane equation can be written using these 3 points using the determinant.
x-x1 y-y1 z-z1
x2-x1 y2-y2 z2-z1
x3-x1 y3-y1 z3-z1 =0
Substitute the values form the points.
We get
![\left[\begin{array}{ccc}x-1&y-3&z-0\\2&-1&1\\0&-2&1\end{array}\right] =0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx-1%26y-3%26z-0%5C%5C2%26-1%261%5C%5C0%26-2%261%5Cend%7Barray%7D%5Cright%5D%20%3D0)
Expand the determinant as
(x-1)(-1+2)-(y-3)(2-0)+z(-4-0) =0
x-1-2y+6-4z =0
x-2y-4z+5 =0