The equation of plane through (–10, 5, 4) is
A(x + 10) + B (y – 5) + C (z – 4) = 0 ...(1)
The direction ratios of the line through the points (4, –1, 2) and (–10.5. 4) are –10 – 4, 5+1, 4-2. i.e.– 14, 6, 2 i.e. 7,–3,–1.
∵ the line with direction ratios 7, –3.–1 'is normal to the plane (1).
∴ equation (1) of plane becomes
or
or
which is the required equation of plane.
Find the equation of a plane which passes through (2, –3, 1) and is perpendicular to the line through the points (3, 4, –1) and (2,–1, 5).