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(A) The required plane passes through the point $(2, -1, 3)$. Let the normal vector to the plane be $\vec{n} = (a, b, c)$. The equation of the plane i

Vedclass · Accepted Answer

$\frac{11}{3}$

Vedclass · Answer

(A) The required plane passes through the point $(2, -1, 3)$. Let the normal vector to the plane be $\vec{n} = (a, b, c)$. The equation of the plane is $a(x - 2) + b(y + 1) + c(z - 3) = 0$. Since the plane is perpendicular to $3x - 2y + z = 9$,the normal vectors are perpendicular,so $3a - 2b + c = 0$. Since the plane is also perpendicular to $x + y + z = 9$,we have $a + b + c = 0$. Solving these equations using cross product: $\frac{a}{(-2)(1) - (1)(1)} = \frac{b}{(1)(1) - (3)(1)} = \frac{c}{(3)(1) - (-2)(1)}$,which gives $\frac{a}{-3} = \frac{b}{-2} = \frac{c}{5} = k$. Thus,the normal vector is proportional to $(-3, -2, 5)$. Substituting these into the plane equation: $-3(x - 2) - 2(y + 1) + 5(z - 3) = 0$. Expanding this: $-3x + 6 - 2y - 2 + 5z - 15 = 0$,which simplifies to $-3x - 2y + 5z - 11 = 0$. Dividing by $-3$ to match the form $x + by + cz + d = 0$: $x + \frac{2}{3}y - \frac{5}{3}z + \frac{11}{3} = 0$. Comparing this with $x + by + cz + d = 0$,we find $d = \frac{11}{3}$.

If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to the planes $3x - 2y + z = 9$ and $x + y + z = 9$ is $x + by + cz + d = 0$,then $d =$

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