The equation of a plane passing through the line of intersection of the planes x + y + z = 5, 2x – y + 3z = 1 and parallel to the line y = z = 0 is:
Answer & explanation
Correct answer: option 2
Plane will be in the form
(x + y + z – 5) + a(2x – y + 3z) –1) = 0 i.e., x(1 + 2a) + y(1 – a) + z(1 + 3a) = 5 + a
It is parallel to the line y = z = 0.
Since, (1 + 2a) = 0
∴ a = $\frac{1}{2}$
Thus required plane is
$\frac{3}{2} y-\frac{1}{2} z=\frac{9}{2}$
i.e., 3y – z = 9