If bar{a}, bar{b}, bar{c} are non-coplanar ve | vedclass

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(C) Let the position vectors of points $A, B$, and $C$ be $\vec{A} = \bar{a}-2 \bar{b}+3 \bar{c}$, $\vec{B} = -4 \bar{a}+5 \bar{b}-6 \bar{c}$, and $\v

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(C) Let the position vectors of points $A, B$, and $C$ be $\vec{A} = \bar{a}-2 \bar{b}+3 \bar{c}$, $\vec{B} = -4 \bar{a}+5 \bar{b}-6 \bar{c}$, and $\vec{C} = x \bar{a}-9 \bar{b}+z \bar{c}$.Since the points $A, B$, and $C$ are collinear, the vector $\vec{AB}$ must be a scalar multiple of $\vec{BC}$ (or $\vec{AC}$).First, calculate $\vec{AB} = \vec{B} - \vec{A} = (-4\bar{a}+5\bar{b}-6\bar{c}) - (\bar{a}-2\bar{b}+3\bar{c}) = -5\bar{a} + 7\bar{b} - 9\bar{c}$.Next, calculate $\vec{BC} = \vec{C} - \vec{B} = (x\bar{a}-9\bar{b}+z\bar{c}) - (-4\bar{a}+5\bar{b}-6\bar{c}) = (x+4)\bar{a} - 14\bar{b} + (z+6)\bar{c}$.Since $A, B, C$ are collinear, $\vec{AB} = k \vec{BC}$ for some scalar $k$.$-5\bar{a} + 7\bar{b} - 9\bar{c} = k((x+4)\bar{a} - 14\bar{b} + (z+6)\bar{c})$.Comparing the coefficients of $\bar{b}$: $7 = -14k \Rightarrow k = -1/2$.Comparing the coefficients of $\bar{a}$: $-5 = k(x+4) \Rightarrow -5 = -1/2(x+4) \Rightarrow 10 = x+4 \Rightarrow x = 6$.Comparing the coefficients of $\bar{c}$: $-9 = k(z+6) \Rightarrow -9 = -1/2(z+6) \Rightarrow 18 = z+6 \Rightarrow z = 12$.Finally, calculate $2x - z = 2(6) - 12 = 12 - 12 = 0$.

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and the points represented by position vectors $\bar{a}-2 \bar{b}+3 \bar{c}$, $-4 \bar{a}+5 \bar{b}-6 \bar{c}$, and $x \bar{a}-9 \bar{b}+z \bar{c}$ are collinear, then $2x-z=$

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