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(D) The equation of a plane passing through the line $\frac{x-1}{1}=\frac{y-2}{5}=\frac{z-3}{4}$ is given by $A(x-1)+B(y-2)+C(z-3)=0$,where $A+5B+4C=0

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$(0, 7, 10)$

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(D) The equation of a plane passing through the line $\frac{x-1}{1}=\frac{y-2}{5}=\frac{z-3}{4}$ is given by $A(x-1)+B(y-2)+C(z-3)=0$,where $A+5B+4C=0$ (since the normal vector is perpendicular to the line direction vector $(1, 5, 4)$).Since the point $(3, 2, 0)$ lies on the plane,we substitute these coordinates into the equation: $A(3-1)+B(2-2)+C(0-3)=0$,which simplifies to $2A-3C=0$,or $2A=3C$.From $A+5B+4C=0$,we substitute $A = \frac{3}{2}C$: $\frac{3}{2}C+5B+4C=0 \Rightarrow 5B = -\frac{11}{2}C \Rightarrow B = -\frac{11}{10}C$.Let $C = -10$,then $A = -15$ and $B = 11$. The equation of the plane is $-15(x-1)+11(y-2)-10(z-3)=0$.$-15x+15+11y-22-10z+30=0 \Rightarrow -15x+11y-10z+23=0$.Checking the point $(0, -3, 1)$: $-15(0)+11(-3)-10(1)+23 = 0 - 33 - 10 + 23 = -20 
eq 0$. Let us re-evaluate the cross product of vectors $(3-1, 2-2, 0-3) = (2, 0, -3)$ and $(1, 5, 4)$.The normal vector $\vec{n} = (2, 0, -3) 	imes (1, 5, 4) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & -3 \ 1 & 5 & 4 \end{vmatrix} = \hat{i}(0 - (-15)) - \hat{j}(8 - (-3)) + \hat{k}(10 - 0) = 15\hat{i} - 11\hat{j} + 10\hat{k}$.The plane equation is $15(x-1) - 11(y-2) + 10(z-3) = 0 \Rightarrow 15x - 11y + 10z - 15 + 22 - 30 = 0 \Rightarrow 15x - 11y + 10z - 23 = 0$.Testing $(0, -3, 1)$: $15(0) - 11(-3) + 10(1) - 23 = 0 + 33 + 10 - 23 = 20 
eq 0$. Testing $(0, 3, 1)$: $15(0) - 11(3) + 10(1) - 23 = -33 + 10 - 23 = -46 
eq 0$. Testing $(0, 7, 10)$: $15(0) - 11(7) + 10(10) - 23 = -77 + 100 - 23 = 0$. Thus,the point is $(0, 7, 10)$.

$A$ plane containing the point $(3, 2, 0)$ and the line $\frac{x - 1}{1} = \frac{y - 2}{5} = \frac{z - 3}{4}$ also contains the point

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