Let the coordinates of the points A and B be (3, – 4, – 5) and (2, – 3, 1) respectively.

The equation of the line joining the points ) is

where r is a constant

Thus, the equation of AB is

Any point on the line AB is the form

– r + 3, r – 4, 6r – 5

Let p be the point of intersection of the line AB and the plane 2x + y + z = 7

Thus, we have,

2(– r + 3) + r – 4 + 6r – 5 = 7

⇒ – 2r + 6 + r – 4 + 6r – 5 = 7

⇒ 5r = 10

⇒ r = 2

Substituting the value of r in – r + 3, r – 4, 6r – 5, the coordinates of P are:

(– 2 + 3, 2 – 4, 12 – 5) = (1, – 2, 7)