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:

3y – z = 9

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