Long Answer Type
The equation of any plane through (0, 0, 0) is
A (x – 0) + B (y – 0) + C (z – 0) = 0
or A x + B y + C z = 0 Â Â Â ...(1)
∴ it passes through (3, –1, 2)
∴ 3A – B + 2C = 0    ...(2)
Since plane (1) is parallel to the line 
∴ normal to the plane with direction ratios A, B, C is perpendicular to the line with direction ratios 1, – 4,  7.
∴ A (1) + B (– 4) + C(7) = 0     [∴ a1 a2 +b1 b2 +c1 c2 = 0]
∴ A – 4B + 7C = 0    ...(3)
Solving (2) and (3), we get,
           
Putting values of A, B, C in (1), we get,
k x – 19 k y – 11 k z = 0
or x – 19 y – 11 z = 0
which is required equation of plane.
Find the coordinates of the image of the point (1, 3, 4) in the plane 2x – y +z + 3 = 0.
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