(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).