The points A(-1, 3, 0),B(2, 2, 1),and C(1, 1, | vedclass

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(A) The equation of a plane passing through $A(-1, 3, 0)$ is given by $A(x + 1) + B(y - 3) + C(z - 0) = 0$ ... $(i)$.Since the plane passes through $B

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$\sqrt{66}$

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(A) The equation of a plane passing through $A(-1, 3, 0)$ is given by $A(x + 1) + B(y - 3) + C(z - 0) = 0$ ... $(i)$.Since the plane passes through $B(2, 2, 1)$ and $C(1, 1, 3)$,we have:$A(2 + 1) + B(2 - 3) + C(1 - 0) = 0 \implies 3A - B + C = 0$ ... $(ii)$.$A(1 + 1) + B(1 - 3) + C(3 - 0) = 0 \implies 2A - 2B + 3C = 0$ ... $(iii)$.Solving $(ii)$ and $(iii)$ using cross-multiplication:$\frac{A}{(-1)(3) - (1)(-2)} = \frac{B}{(1)(2) - (3)(3)} = \frac{C}{(3)(-2) - (-1)(2)}$$\frac{A}{-3 + 2} = \frac{B}{2 - 9} = \frac{C}{-6 + 2} \implies \frac{A}{-1} = \frac{B}{-7} = \frac{C}{-4}$.Thus,the direction ratios are $(1, 7, 4)$.Substituting into $(i)$: $1(x + 1) + 7(y - 3) + 4(z) = 0 \implies x + 7y + 4z - 20 = 0$.The distance from the point $D(5, 7, 8)$ to the plane $x + 7y + 4z - 20 = 0$ is:$d = \frac{|1(5) + 7(7) + 4(8) - 20|}{\sqrt{1^2 + 7^2 + 4^2}} = \frac{|5 + 49 + 32 - 20|}{\sqrt{1 + 49 + 16}} = \frac{66}{\sqrt{66}} = \sqrt{66}$.

The points $A(-1, 3, 0)$,$B(2, 2, 1)$,and $C(1, 1, 3)$ determine a plane. The distance from the plane to the point $D(5, 7, 8)$ is

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