If A(2,1,-1), B(6,-3,2), and C(-3,12,4) are t | vedclass

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(D) The normal vector $\vec{n}$ to the plane is given by the cross product of vectors $\vec{AB}$ and $\vec{AC}$.$\vec{AB} = (6-2, -3-1, 2-(-1)) = (4,

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$-15$

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(D) The normal vector $\vec{n}$ to the plane is given by the cross product of vectors $\vec{AB}$ and $\vec{AC}$.$\vec{AB} = (6-2, -3-1, 2-(-1)) = (4, -4, 3)$.$\vec{AC} = (-3-2, 12-1, 4-(-1)) = (-5, 11, 5)$.$\vec{n} = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & -4 & 3 \ -5 & 11 & 5 \end{vmatrix} = \hat{i}(-20-33) - \hat{j}(20+15) + \hat{k}(44-20) = -53\hat{i} - 35\hat{j} + 24\hat{k}$.The equation of the plane is $-53(x-2) - 35(y-1) + 24(z+1) = 0$.$-53x + 106 - 35y + 35 + 24z + 24 = 0$.$-53x - 35y + 24z + 165 = 0$.Multiply by $-1$ to match the form $53x + by + cz + d = 0$:$53x + 35y - 24z - 165 = 0$.Here, $b = 35$, $c = -24$, and $d = -165$.Then, $\frac{d}{b+c} = \frac{-165}{35 - 24} = \frac{-165}{11} = -15$.

If $A(2,1,-1)$, $B(6,-3,2)$, and $C(-3,12,4)$ are the vertices of a triangle $ABC$, and the equation of the plane containing the triangle $ABC$ is $53x + by + cz + d = 0$, then find the value of $\frac{d}{b+c}$.

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