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(D) The equation of a plane passing through a point $(x_1, y_1, z_1)$ is given by $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$. Substituting the point $(0, 1,

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

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(D) The equation of a plane passing through a point $(x_1, y_1, z_1)$ is given by $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$. Substituting the point $(0, 1, 2)$,we get $A(x-0) + B(y-1) + C(z-2) = 0$ ... $(i)$. Since the plane passes through $(-1, 0, 3)$,we have $A(-1-0) + B(0-1) + C(3-2) = 0$,which simplifies to $-A - B + C = 0$ or $A + B - C = 0$ ... $(ii)$. The plane $(i)$ is perpendicular to $2x + 3y + z = 5$,so the normal vectors are perpendicular,giving $2A + 3B + C = 0$ ... $(iii)$. Solving $(ii)$ and $(iii)$ using cross multiplication: $\frac{A}{1(1) - 3(-1)} = \frac{B}{2(-1) - 1(1)} = \frac{C}{1(3) - 2(1)}$. This gives $\frac{A}{4} = \frac{B}{-3} = \frac{C}{1}$. Substituting these values into $(i)$,we get $4(x-0) - 3(y-1) + 1(z-2) = 0$,which simplifies to $4x - 3y + z + 1 = 0$.

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

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