The equation of the line through (3, –4, –5) and (2, –3, 1) is $\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}$ ...(1) Any point on (1) is (3 – k, – 4 + k, – 5 + 6k) ...(2)This point lies on 2x + y + z = 7∴ 2 (3 – k) + (– 4 + k) + (–5 + 6k) = 7⇒ 6 – 2k – 4 + k – 5 + 6k = 7 ⇒ 5k = 10 ⇒ k = 2Putting in (2), the required point is (3 –2, – 4 + 2, – 5 + 12) i.e. (1, – 2, 7)