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  4. A plane which passes through the point (3, 2, 0) and the line (x-4/1) = (y-7/5) = (z-4/4) is

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Q. A plane which passes through the point (3, 2, 0) and the line $\frac{x-4}{1} = \frac{y-7}{5} = \frac{z-4}{4} $ is

Three Dimensional Geometry

Solution:

As the point (3, 2, 0) lies on the given line $\frac{x-4}{1} = \frac{y-7}{5} = \frac{z-4}{4} $
$\therefore $ There can be infinite many planes passing through this line. But here out of the four options only first option is satisfied by the coordinates of both the points (3, 2, 0) and (4, 7, 4)
$\therefore $ x - y + z = 1 is the required plane.