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

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$3x - 2y - z = 4$

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(C) The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$. Here,the planes are $2x - y - 4 = 0$ and $y + 2z - 4 = 0$. So,the equation of the required plane is $(2x - y - 4) + \lambda(y + 2z - 4) = 0$  --- $(i)$. Since the plane passes through the point $(2, 1, 0)$,we substitute $x = 2, y = 1, z = 0$ into equation $(i)$: $(2(2) - 1 - 4) + \lambda(1 + 2(0) - 4) = 0$ $(4 - 1 - 4) + \lambda(1 - 4) = 0$ $-1 - 3\lambda = 0$ $-3\lambda = 1$ $\lambda = -\frac{1}{3}$. Substituting $\lambda = -\frac{1}{3}$ back into equation $(i)$: $(2x - y - 4) - \frac{1}{3}(y + 2z - 4) = 0$ Multiply the entire equation by $3$: $3(2x - y - 4) - (y + 2z - 4) = 0$ $6x - 3y - 12 - y - 2z + 4 = 0$ $6x - 4y - 2z - 8 = 0$ Dividing by $2$: $3x - 2y - z - 4 = 0$ $3x - 2y - z = 4$.

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 $(2, 1, 0)$,is

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