BITSAT 2005 Solved Question Paper

Given equation of line is

x−1

3

=

y+2

4

=

z−3

−2

(say)
Any point on the line is (3k+1,4k−2−2k+3)
If the given line intersect the plane 2x−y+3z−1=0, then any point on the line lies in the plane.
∴(3k+1)−(4k−2)+3(−2k +3)−1=0
⇒−4k+12=0⇒k=3
∴ Point is (9+1,12−2,−6+3)
i.e., (10,10,−3).