If $[(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}) \times(\overline{\mathrm{b}}+2…

If $[(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}) \times(\overline{\mathrm{b}}+2 \overline{\mathrm{c}}+3 \overline{\mathrm{a}})] \cdot(\overline{\mathrm{c}}+2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}})=54$, then the value of $\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$ is
  1. 0
  2. 1
  3. 3
  4. 2

Solution

R.H.S. of the given equality can be written as $\begin{aligned} &(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot[(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}) \times(3 \overline{\mathrm{a}}+\overline{\mathrm{b}}+2 \overline{\mathrm{c}})] \\ &=(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot[3(\overline{\mathrm{a}} \times \overline{\mathrm{a}})+(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{a}} \times \overline{\mathrm{c}}) \\ &+6(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{b}})+4(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\ &+9(\overline{\mathrm{c}} \times \overline{\mathrm{a}})+3(\overline{\mathrm{c}} \times \overline{\mathrm{b}})+6(\overline{\mathrm{c}} \times \overline{\mathrm{c}})] \\ &=(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}})[0+(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{a}} \times \overline{\mathrm{c}}) \\ &-6(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+0+4(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\ &=(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}})[-5(\overline{\mathrm{a}} \times \overline{\mathrm{c}})-3(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+(\overline{\mathrm{b}} \times \overline{\mathrm{c}})-7(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \\ &=-10[\overline{\mathrm{a}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})]+2[\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]-14[\overline{\mathrm{a}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \\ &-15[\overline{\mathrm{b}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})]+3[\overline{\mathrm{b}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]-21[\overline{\mathrm{b}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \\ &-5[\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})]+[\overline{\mathrm{c}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]-7[\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \\ &= 0+2[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+0 \\ &+0+0 \\ &=-5[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+0+21[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}] \\ &= 18[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=54 \end{aligned}$ $\Rightarrow[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=3$

Asked in: MHT CET 2023 (09 May Shift 1)

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