The equation of the plane passing through (1, | vedclass

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(A) The equation of a plane passing through $(1, -1, 2)$ is given by $a(x - 1) + b(y + 1) + c(z - 2) = 0$. Since this plane is perpendicular to the pl

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$6x - 7y - 4z - 5 = 0$

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(A) The equation of a plane passing through $(1, -1, 2)$ is given by $a(x - 1) + b(y + 1) + c(z - 2) = 0$. Since this plane is perpendicular to the planes $x + 2y - 2z = 4$ and $3x + 2y + z = 6$,its normal vector $(a, b, c)$ must be perpendicular to the normal vectors of the given planes,which are $\vec{n_1} = (1, 2, -2)$ and $\vec{n_2} = (3, 2, 1)$. Thus,we have the equations: $a + 2b - 2c = 0$ $3a + 2b + c = 0$ Using the cross product to find the direction ratios $(a, b, c)$: $a = (2)(1) - (-2)(2) = 2 + 4 = 6$ $b = (-2)(3) - (1)(1) = -6 - 1 = -7$ $c = (1)(2) - (2)(3) = 2 - 6 = -4$ So,the direction ratios are $(6, -7, -4)$. Substituting these into the plane equation: $6(x - 1) - 7(y + 1) - 4(z - 2) = 0$ $6x - 6 - 7y - 7 - 4z + 8 = 0$ $6x - 7y - 4z - 5 = 0$.

The equation of the plane passing through $(1, -1, 2)$ and perpendicular to the planes $x + 2y - 2z = 4$ and $3x + 2y + z = 6$ is:

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