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(B) Let the position vectors of the four points be $\vec{P} = 2\bar{a}+3\bar{b}-\bar{c}$, $\vec{Q} = \bar{a}-2\bar{b}+3\bar{c}$, $\vec{R} = 3\bar{a}+4

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Coplanar

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(B) Let the position vectors of the four points be $\vec{P} = 2\bar{a}+3\bar{b}-\bar{c}$, $\vec{Q} = \bar{a}-2\bar{b}+3\bar{c}$, $\vec{R} = 3\bar{a}+4\bar{b}-2\bar{c}$, and $\vec{S} = \bar{a}-6\bar{b}+6\bar{c}$.To check if these points are coplanar, we examine the vectors $\vec{PQ}$, $\vec{PR}$, and $\vec{PS}$.$\vec{PQ} = \vec{Q} - \vec{P} = (\bar{a}-2\bar{b}+3\bar{c}) - (2\bar{a}+3\bar{b}-\bar{c}) = -\bar{a}-5\bar{b}+4\bar{c}$.$\vec{PR} = \vec{R} - \vec{P} = (3\bar{a}+4\bar{b}-2\bar{c}) - (2\bar{a}+3\bar{b}-\bar{c}) = \bar{a}+\bar{b}-\bar{c}$.$\vec{PS} = \vec{S} - \vec{P} = (\bar{a}-6\bar{b}+6\bar{c}) - (2\bar{a}+3\bar{b}-\bar{c}) = -\bar{a}-9\bar{b}+7\bar{c}$.Four points are coplanar if the scalar triple product $[\vec{PQ}, \vec{PR}, \vec{PS}] = 0$.Calculating the determinant of the coefficients of $\bar{a}, \bar{b}, \bar{c}$:$D = \begin{vmatrix} -1 & -5 & 4 \ 1 & 1 & -1 \ -1 & -9 & 7 \end{vmatrix}$.$D = -1(7 - 9) - (-5)(7 - 1) + 4(-9 + 1) = -1(-2) + 5(6) + 4(-8) = 2 + 30 - 32 = 0$.Since the scalar triple product is $0$, the four points are coplanar.

The four points whose position vectors are given by $2\bar{a}+3\bar{b}-\bar{c}$, $\bar{a}-2\bar{b}+3\bar{c}$, $3\bar{a}+4\bar{b}-2\bar{c}$ and $\bar{a}-6\bar{b}+6\bar{c}$ are

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