If x and y are real numbers such that hat{i}+ | vedclass

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(B) Let the four points be $A(1, 1, 1)$, $B(-2, 3, 2)$, $C(x, -5, 3)$, and $D(1, y, -1)$.Since the points are coplanar, the vectors $\vec{AB}$, $\vec{

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$(x+5)(y+3)=60$

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(B) Let the four points be $A(1, 1, 1)$, $B(-2, 3, 2)$, $C(x, -5, 3)$, and $D(1, y, -1)$.Since the points are coplanar, the vectors $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$ must be coplanar, which implies their scalar triple product is zero:$\vec{AB} = (-2-1)\hat{i} + (3-1)\hat{j} + (2-1)\hat{k} = -3\hat{i} + 2\hat{j} + \hat{k}$$\vec{AC} = (x-1)\hat{i} + (-5-1)\hat{j} + (3-1)\hat{k} = (x-1)\hat{i} - 6\hat{j} + 2\hat{k}$$\vec{AD} = (1-1)\hat{i} + (y-1)\hat{j} + (-1-1)\hat{k} = 0\hat{i} + (y-1)\hat{j} - 2\hat{k}$Setting the determinant to zero:$\begin{vmatrix} -3 & 2 & 1 \ x-1 & -6 & 2 \ 0 & y-1 & -2 \end{vmatrix} = 0$Expanding along the first column:$-3[(-6)(-2) - 2(y-1)] - (x-1)[(2)(-2) - (1)(y-1)] + 0 = 0$$-3[12 - 2y + 2] - (x-1)[-4 - y + 1] = 0$$-3[14 - 2y] - (x-1)[-y - 3] = 0$$-42 + 6y + (x-1)(y+3) = 0$$-42 + 6y + xy + 3x - y - 3 = 0$$xy + 3x + 5y - 45 = 0$Adding $15$ to both sides to factorize:$xy + 3x + 5y + 15 = 45 + 15$$x(y+3) + 5(y+3) = 60$$(x+5)(y+3) = 60$

If $x$ and $y$ are real numbers such that $\hat{i}+\hat{j}+\hat{k}$, $-2 \hat{i}+3 \hat{j}+2 \hat{k}$, $x \hat{i}-5 \hat{j}+3 \hat{k}$, and $\hat{i}+y \hat{j}-\hat{k}$ are the position vectors of four coplanar points, then the locus of $P(x, y)$ is

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