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(A) The equation of a plane passing through the line of intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$. Given

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$x - y - z = 0$

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(A) The equation of a plane passing through the line of intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$. Given planes are $2x - y - 4 = 0$ and $y + 2z - 4 = 0$. So,the equation of the required plane is $(2x - y - 4) + \lambda(y + 2z - 4) = 0$. Since the plane passes through the point $(1, 1, 0)$,we substitute $x = 1, y = 1, z = 0$ into the equation: $(2(1) - 1 - 4) + \lambda(1 + 2(0) - 4) = 0$ $(2 - 1 - 4) + \lambda(1 - 4) = 0$ $-3 - 3\lambda = 0$ $-3\lambda = 3$ $\lambda = -1$. Substituting $\lambda = -1$ back into the equation: $(2x - y - 4) - 1(y + 2z - 4) = 0$ $2x - y - 4 - y - 2z + 4 = 0$ $2x - 2y - 2z = 0$ Dividing by $2$,we get $x - y - z = 0$.

The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1, 1, 0)$ is:

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