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(D) The equation of a plane passing through $(0, 1, 2)$ is given by $a(x - 0) + b(y - 1) + c(z - 2) = 0$,which simplifies to $ax + b(y - 1) + c(z - 2)

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$4x - 3y + z + 1 = 0$

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(D) The equation of a plane passing through $(0, 1, 2)$ is given by $a(x - 0) + b(y - 1) + c(z - 2) = 0$,which simplifies to $ax + b(y - 1) + c(z - 2) = 0 \dots (i)$.Since the plane passes through $(-1, 0, 3)$,we substitute these coordinates into $(i)$:$a(-1 - 0) + b(0 - 1) + c(3 - 2) = 0 \implies -a - b + c = 0 \implies a + b - c = 0 \dots (ii)$.The plane $(i)$ is perpendicular to the plane $2x + 3y + z = 5$,so the normal vectors are perpendicular,implying $2a + 3b + c = 0 \dots (iii)$.Solving $(ii)$ and $(iii)$ using cross multiplication:$\frac{a}{1(1) - (-1)(3)} = \frac{-b}{1(1) - (-1)(2)} = \frac{c}{1(3) - 1(2)}$$\frac{a}{1 + 3} = \frac{-b}{1 + 2} = \frac{c}{3 - 2} \implies \frac{a}{4} = \frac{-b}{3} = \frac{c}{1} = k$.Thus,$a = 4k, b = -3k, c = k$.Substituting these into $(i)$:$4k(x) - 3k(y - 1) + k(z - 2) = 0$.Dividing by $k$ $(k 
eq 0)$:$4x - 3y + 3 + z - 2 = 0 \implies 4x - 3y + z + 1 = 0$.

The equation of the plane passing through the points $(0, 1, 2)$ and $(-1, 0, 3)$ and perpendicular to the plane $2x + 3y + z = 5$ is

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