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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,$(2x - 7y

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$4$

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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,$(2x - 7y + 4z - 3) + \lambda(3x - 5y + 4z + 11) = 0$.Grouping the terms,we get $(2 + 3\lambda)x - (7 + 5\lambda)y + (4 + 4\lambda)z + (-3 + 11\lambda) = 0$.Since the plane passes through the point $(-2, 1, 3)$,we substitute these coordinates into the equation:$(2 + 3\lambda)(-2) - (7 + 5\lambda)(1) + (4 + 4\lambda)(3) - 3 + 11\lambda = 0$.$-4 - 6\lambda - 7 - 5\lambda + 12 + 12\lambda - 3 + 11\lambda = 0$.Combining the $\lambda$ terms and constants: $(-6 - 5 + 12 + 11)\lambda + (-4 - 7 + 12 - 3) = 0$.$12\lambda - 2 = 0$,which gives $\lambda = \frac{1}{6}$.Substituting $\lambda = \frac{1}{6}$ back into the plane equation:$(2 + 3(\frac{1}{6}))x - (7 + 5(\frac{1}{6}))y + (4 + 4(\frac{1}{6}))z + (-3 + 11(\frac{1}{6})) = 0$.$(\frac{15}{6})x - (\frac{47}{6})y + (\frac{28}{6})z - (\frac{7}{6}) = 0$.Multiplying by $6$,we get $15x - 47y + 28z - 7 = 0$.Comparing this with $ax + by + cz - 7 = 0$,we have $a = 15, b = -47, c = 28$.Now,calculate $2a + b + c - 7 = 2(15) + (-47) + 28 - 7 = 30 - 47 + 28 - 7 = 4$.

If the equation of the plane passing through the line of intersection of the planes $2x - 7y + 4z - 3 = 0$ and $3x - 5y + 4z + 11 = 0$ and the point $(-2, 1, 3)$ is $ax + by + cz - 7 = 0$,then the value of $2a + b + c - 7$ is

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