The point of intersection of the line `(x-1)/3=(y+2)/4=(z-3)/-2` and the plane `2x-y+3z-1=0`, is

Options

A. `(-10,10,3)`
B. `(10,10,-3)`
C. `(10,-10,3)`
D. `(10,-10,-3)`

Answer

b

Explanation

The given line is `(x-1)/3=(y+2)/4=(z-3)/-2=lambda` say.
A general point on this line is `(3lambda+1,4lambda-2,-2lambda+3)`.
For some value of `lambda`, let the point `P(3lambda+1,4lambda-2,-2lambda+3)` lie on the plane `2x-y+3z-1=0`. Then,
`2(3lambda+1)-(4lambbda-2)+3(-2lambda+3)-1=0`.
`rArr 4lambda=12 rArr lambda=3`.
So, the required point is P(9+1, 12-2,-6+3)`. i.e, P(10,10,-3)`.