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The equation of the plane passing through the point (1, 2, - 3) and perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7, is :

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6x – 5y + 2z + 10 = 0

3x – 10y - 2z + 11 = 0

11x + y + 17z + 38 = 0

6x – 5y - 2z - 2 = 0

answer is C.

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Detailed Solution

Normal vector of required plane is $\vec{n} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & - 2 \\ 2 & - 5 & - 1 \end{matrix}| = - 11 \hat{i} - \hat{j} - 17 \hat{k}$

Therefore, the equation of the plane is 11 (x – 1) + (y – 2) + 17 (z + 3) = 0, it implies 11x + y + 17z + 38 = 0

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