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The line (x-1)/2=(y-2)/3=(z-3)/4 meets t...

The line `(x-1)/2=(y-2)/3=(z-3)/4` meets the plane `2x+3y-z=14` in the point

A

(2,5,7)

B

(3,5,7)

C

(5,7,3)

D

(6,5,3)

Text Solution

Verified by Experts

The correct Answer is:
b
Let `(x-1)/2=(y-2)/3=(z-3)/4 = lambda` (say).
A general point on this lines is `(2lambda+1, 3lambda+2, 4lambda+3)`.
For some value of `lambda`, let the given line meet the plane `2x+3y-z=14` at a point `P(2lambda+1,3lambda+2, 4lambda+3)`. Then,
`2(2lambda+1)+3(3lambda+2)-(4lambda+3)=14 rArr 9lambda=9 rArr lambda=1`.
So, the required points is `P(2+1,3+2,4+3)`, i.e., P(3,5,7).
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