Given – A plane passes through (2, - 1, 5), perpendicular to the plane x + 2y - 3z = 7 and parallel to the line

To find – The equation of the plane

Let the equation of the required plane be ax + by + cz + d = 0……(a)

The plane passes through (2, - 1, 5)

So, 2a - b + 5c + d = 0…………………(i)

The direction ratios of the normal of the plane is given by (a, b, c)

Now, this plane is perpendicular to the plane x + 2y - 3z = 7 having direction ratios (1, 2, - 3)

So, a + 2b - 3c = 0……………(ii)

This plane is also parallel to the line having direction ratios (3, - 1, 1)

So, the direction of the normal of the required plane is also at right angles to the given line.

So, 3a - b + c = 0………………(iii)

Solving equations (ii) and (iii),

[α → arbitrary constant]

Putting these values in equation (i) we get,

2Х(- α) - (- 10α) + 5(- 7α) + d = 0 i.e. d = 27α

Substituting all the values of a, b, c and d in equation (a) we get,