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(D) The equation of a plane passing through $(1, 0, -1)$ is given by $a(x - 1) + b(y - 0) + c(z + 1) = 0 \dots (i)$.Since the plane passes through $(3

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(D) The equation of a plane passing through $(1, 0, -1)$ is given by $a(x - 1) + b(y - 0) + c(z + 1) = 0 \dots (i)$.Since the plane passes through $(3, 2, 2)$,we have $a(3 - 1) + b(2 - 0) + c(2 + 1) = 0$,which simplifies to $2a + 2b + 3c = 0 \dots (ii)$.The plane is parallel to the line with direction ratios $(2, -2, 3)$,so the normal to the plane is perpendicular to the line. Thus,$2a - 2b + 3c = 0 \dots (iii)$.Subtracting $(iii)$ from $(ii)$,we get $4b = 0$,so $b = 0$.Substituting $b = 0$ into $(ii)$,we get $2a + 3c = 0$,or $a = -\frac{3}{2}c$. Let $c = 2$,then $a = -3$.The normal vector is $(-3, 0, 2)$.The equation of the plane is $-3(x - 1) + 0(y - 0) + 2(z + 1) = 0$,which simplifies to $-3x + 3 + 2z + 2 = 0$,or $3x - 2z - 5 = 0$.

The equation of the plane passing through the points $(3, 2, 2)$ and $(1, 0, -1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y - 1}{-2} = \frac{z - 2}{3}$ is:

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