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(C) The general equation of a plane passing through $A(2, 2, -1)$ is given by:$a(x - 2) + b(y - 2) + c(z + 1) = 0 \dots(i)$Since the plane passes thro

Vedclass · Accepted Answer

$5x + 2y - 3z = 17$

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(C) The general equation of a plane passing through $A(2, 2, -1)$ is given by:$a(x - 2) + b(y - 2) + c(z + 1) = 0 \dots(i)$Since the plane passes through $B(3, 4, 2)$ and $C(7, 0, 6)$,we substitute these points into $(i)$:For $B(3, 4, 2)$: $a(3 - 2) + b(4 - 2) + c(2 + 1) = 0 \implies a + 2b + 3c = 0 \dots(ii)$For $C(7, 0, 6)$: $a(7 - 2) + b(0 - 2) + c(6 + 1) = 0 \implies 5a - 2b + 7c = 0 \dots(iii)$Solving $(ii)$ and $(iii)$ using cross-multiplication:$\frac{a}{(2)(7) - (3)(-2)} = \frac{b}{(3)(5) - (1)(7)} = \frac{c}{(1)(-2) - (2)(5)}$$\frac{a}{14 + 6} = \frac{b}{15 - 7} = \frac{c}{-2 - 10}$$\frac{a}{20} = \frac{b}{8} = \frac{c}{-12} \implies \frac{a}{5} = \frac{b}{2} = \frac{c}{-3} = k$Thus,$a = 5k, b = 2k, c = -3k$. Substituting these into $(i)$:$5k(x - 2) + 2k(y - 2) - 3k(z + 1) = 0$$5(x - 2) + 2(y - 2) - 3(z + 1) = 0$$5x - 10 + 2y - 4 - 3z - 3 = 0$$5x + 2y - 3z = 17$.

Find the equation of the plane passing through the points $A(2, 2, -1)$,$B(3, 4, 2)$,and $C(7, 0, 6)$.

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