Vectors $\vec{p}=a \hat{i}+b \hat{j}+c \hat{k}, \vec{q}=d \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}$ forming a triangle $\mathrm{ABC}$ are such that $\vec{p}=\vec{q}+\vec{r}$. If the area of $\triangle \mathrm{ABC}$ is $5 \sqrt{6}$ sq. units, then the sum of the absolute values of $a, b, c$ is

• Correct answer
  A

  $14$
• B

  $13$
• C

  $12$
• D

  $10$

Solution

$a \hat{i}+b \hat{j}+c \hat{k}=d \hat{i}+3 \hat{j}+4 \hat{k}+3 \hat{i}+\hat{j}-2 \hat{k}$ $\begin{aligned} & a \hat{i}+b \hat{j}+c \hat{k}=(d+3) \hat{i}+4 \hat{j}+2 \hat{k} \\ & \therefore \quad a=d+3 \Rightarrow d=a-3 \\ & \quad b=4 \\ & \quad c=2\end{aligned}$ Area of $\begin{aligned} \Delta & =\frac{1}{2}|\vec{p} \times \vec{q}| \\ & =\frac{1}{2}\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ a & b & c \\ d & 3 & 4\end{array}\right|=5 \sqrt{6} .\end{aligned}$ $\begin{aligned} & \Rightarrow|(4 b-3 c) \hat{i}+(c d-4 a) \hat{j}+(3 a-b d) \hat{k}|=10 \sqrt{6} \\ & \Rightarrow|10 \hat{i}+(2 d-4 a) \hat{j}+(3 a-4 d) \hat{k}|=10 \sqrt{6} \\ & \Rightarrow|10 \hat{i}+(-2 a-6) \hat{j}+(12-a) \hat{k}|=10 \sqrt{6} \\ & \Rightarrow \quad 10^2+(-2 a-6)^2+(12-a)^2=600 \\ & \Rightarrow \quad 4 a^2+36+24 a+144+a^2-24 a=500 \\ & \Rightarrow 5 a^2=320 \Rightarrow a^2=64 \\ & \therefore|a|=8 \\ & \therefore|a|+|b|+|c|=8+4+2=14 .\end{aligned}$

Hence, the correct answer is option: 'A'.