Given p = 2a - 3b,q = a - 2b + c,and r = -3a | vedclass

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(B) Let $-2a + 3b - c = xp + yq + zr$. Substituting the expressions for $p, q,$ and $r$: $-2a + 3b - c = x(2a - 3b) + y(a - 2b + c) + z(-3a + b + 2c)$

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$\frac{-7q + r}{5}$

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(B) Let $-2a + 3b - c = xp + yq + zr$. Substituting the expressions for $p, q,$ and $r$: $-2a + 3b - c = x(2a - 3b) + y(a - 2b + c) + z(-3a + b + 2c)$ $-2a + 3b - c = (2x + y - 3z)a + (-3x - 2y + z)b + (y + 2z)c$. Comparing the coefficients of $a, b,$ and $c$: $1) 2x + y - 3z = -2$ $2) -3x - 2y + z = 3$ $3) y + 2z = -1$ From $(3)$,$y = -1 - 2z$. Substituting into $(1)$ and $(2)$: $2x + (-1 - 2z) - 3z = -2 \implies 2x - 5z = -1$ $-3x - 2(-1 - 2z) + z = 3 \implies -3x + 2 + 4z + z = 3 \implies -3x + 5z = 1$ Adding the two equations: $(2x - 5z) + (-3x + 5z) = -1 + 1 \implies -x = 0 \implies x = 0$. Then $5z = 1 \implies z = \frac{1}{5}$. $y = -1 - 2(\frac{1}{5}) = -1 - \frac{2}{5} = -\frac{7}{5}$. Thus,$-2a + 3b - c = 0p - \frac{7}{5}q + \frac{1}{5}r = \frac{-7q + r}{5}$.

Given $p = 2a - 3b$,$q = a - 2b + c$,and $r = -3a + b + 2c$,where $a, b,$ and $c$ are non-zero,non-coplanar vectors,then the vector $-2a + 3b - c$ is equal to:

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