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(B) The equation of any plane passing through the intersection of two planes $P_1$ and $P_2$ is given by $P_1 + \lambda P_2 = 0$. Substituting the giv

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$x - 3y + 2z + 1 = 0$

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(B) The equation of any plane passing through the intersection of two planes $P_1$ and $P_2$ is given by $P_1 + \lambda P_2 = 0$. Substituting the given equations: $(2x - y + z - 2) + \lambda (x + 2y - z - 3) = 0$. Since the plane passes through the point $(3, 2, 1)$,we substitute $x=3, y=2, z=1$ into the equation: $(2(3) - 2 + 1 - 2) + \lambda (3 + 2(2) - 1 - 3) = 0$. $(6 - 2 + 1 - 2) + \lambda (3 + 4 - 1 - 3) = 0$. $3 + \lambda (3) = 0$. $3\lambda = -3$,which gives $\lambda = -1$. Substituting $\lambda = -1$ back into the equation: $(2x - y + z - 2) - 1(x + 2y - z - 3) = 0$. $2x - x - y - 2y + z + z - 2 + 3 = 0$. $x - 3y + 2z + 1 = 0$.

Let two planes be given by $P_1 : 2x - y + z = 2$ and $P_2 : x + 2y - z = 3$. Based on the given information,find the equation of the plane passing through the intersection of $P_1$ and $P_2$ and the point $(3, 2, 1)$.

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