(a-2 b-c) [(a-b) a-b-c]=3[a-b-c] Question is modified. (a-2 b-c)[(a-b) a-b-c]=3[a b c]

& (a+2 b-c) [(a-b) (a-b-c)] & =(a+2 b-c) (a a-a b-a c-b a+b b+b c & =(a+2 b-c) (0-a b+c a+a b+0+b c) & =(a+2 b-c) (c a+b c) =a (c a)+a (b c)+2 b (c a)+2 b (b c)- c (c a)-c (b c) =0+a (b c)+2 b (c a)+2 0-0-0 & = [ lll a & b & c ]+2 [ lll b & c & a ] & = [ lll a & b & c ]+2 [ lll a & b & c ]=3 [ lll a & b & c ]

(a-2 b-c) [(a-b) a-b-c]=3[a-b-c] Question is modified. (a-2 b-c)[…

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Question

$(\bar{a}-2 \bar{b}-\bar{c}) \cdot[(\bar{a}-\bar{b}) \times \bar{a}-\bar{b}-\bar{c}]=3[\bar{a}-\bar{b}-\bar{c}]$

Question is modified.

$(\bar{a}-2 \bar{b}-\bar{c})[(\bar{a}-\bar{b}) \times \bar{a}-\bar{b}-\bar{c}]=3[\bar{a} \bar{b} \bar{c}]$

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Answer

$\begin{aligned} & (\bar{a}+2 \bar{b}-\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})] \\ & =(\bar{a}+2 \bar{b}-\bar{c}) \cdot(\bar{a} \times \bar{a}-\bar{a} \times \bar{b}-\bar{a} \times \bar{c}-\bar{b} \times \bar{a}+\bar{b} \times \bar{b}+\bar{b} \times \bar{c} \\ & =(\bar{a}+2 \bar{b}-\bar{c}) \cdot(\overline{0}-\bar{a} \times \bar{b}+\bar{c} \times \bar{a}+\bar{a} \times \bar{b}+\overline{0}+\bar{b} \times \bar{c}) \\ & =(\bar{a}+2 \bar{b}-\bar{c}) \cdot(\bar{c} \times \bar{a}+\bar{b} \times \bar{c})\end{aligned}$

$=\bar{a} \cdot(\bar{c} \times \bar{a})+\bar{a} \cdot(\bar{b} \times \bar{c})+2 \bar{b} \cdot(\bar{c} \times \bar{a})+2 \bar{b} \cdot(\bar{b} \times \bar{c})-$

$\bar{c} \cdot(\bar{c} \times \bar{a})-\bar{c} \cdot(\bar{b} \times \bar{c})$

$=0+\bar{a} \cdot(\bar{b} \times \bar{c})+2 \bar{b} \cdot(\bar{c} \times \bar{a})+2 \times 0-0-0$

$\begin{aligned} & =\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]+2\left[\begin{array}{lll}\bar{b} & \bar{c} & \bar{a}\end{array}\right] \\ & =\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]+2\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]=3\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]\end{aligned}$