If the points A(1,1,2), B(2,1, p), C(1,0,3) a | vedclass

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(C) Four points $A, B, C,$ and $D$ are coplanar if the scalar triple product of vectors $\vec{AB}, \vec{AC},$ and $\vec{AD}$ is zero,i.e.,$[\vec{AB},

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(C) Four points $A, B, C,$ and $D$ are coplanar if the scalar triple product of vectors $\vec{AB}, \vec{AC},$ and $\vec{AD}$ is zero,i.e.,$[\vec{AB}, \vec{AC}, \vec{AD}] = 0$.First,we find the vectors:$\vec{AB} = (2-1)\hat{i} + (1-1)\hat{j} + (p-2)\hat{k} = \hat{i} + 0\hat{j} + (p-2)\hat{k}$$\vec{AC} = (1-1)\hat{i} + (0-1)\hat{j} + (3-2)\hat{k} = 0\hat{i} - \hat{j} + \hat{k}$$\vec{AD} = (2-1)\hat{i} + (2-1)\hat{j} + (0-2)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$Now,calculate the determinant:$\begin{vmatrix} 1 & 0 & p-2 \ 0 & -1 & 1 \ 1 & 1 & -2 \end{vmatrix} = 0$Expanding along the first row:$1((-1)(-2) - (1)(1)) - 0(...) + (p-2)((0)(1) - (-1)(1)) = 0$$1(2 - 1) + (p-2)(1) = 0$$1 + p - 2 = 0$$p - 1 = 0$$p = 1$

If the points $A(1,1,2), B(2,1, p), C(1,0,3)$ and $D(2,2,0)$ are coplanar,then the value of $p$ is

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