Vectors vec{p}=a hat{i}+b hat{j}+c hat{k},vec | vedclass

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(A) Given $\vec{p} = \vec{q} + \vec{r}$,we have:$a \hat{i} + b \hat{j} + c \hat{k} = (d \hat{i} + 3 \hat{j} + 4 \hat{k}) + (3 \hat{i} + \hat{j} - 2 \h

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$14$

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(A) Given $\vec{p} = \vec{q} + \vec{r}$,we have:$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})$$a \hat{i} + b \hat{j} + c \hat{k} = (d + 3) \hat{i} + 4 \hat{j} + 2 \hat{k}$Comparing coefficients,we get $a = d + 3 \Rightarrow d = a - 3$,$b = 4$,and $c = 2$.The area of $	riangle ABC$ formed by vectors $\vec{q}$ and $\vec{r}$ is given by $\frac{1}{2} |\vec{q} 	imes \vec{r}| = 5 \sqrt{6}$.$\vec{q} 	imes \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ d & 3 & 4 \ 3 & 1 & -2 \end{vmatrix} = \hat{i}(-6 - 4) - \hat{j}(-2d - 12) + \hat{k}(d - 9) = -10 \hat{i} + (2d + 12) \hat{j} + (d - 9) \hat{k}$.Area $= \frac{1}{2} \sqrt{(-10)^2 + (2d + 12)^2 + (d - 9)^2} = 5 \sqrt{6}$.$\sqrt{100 + 4d^2 + 48d + 144 + d^2 - 18d + 81} = 10 \sqrt{6}$.$5d^2 + 30d + 325 = 600 \Rightarrow 5d^2 + 30d - 275 = 0 \Rightarrow d^2 + 6d - 55 = 0$.$(d + 11)(d - 5) = 0$,so $d = 5$ or $d = -11$.If $d = 5$,$a = 5 + 3 = 8$. Then $|a| + |b| + |c| = 8 + 4 + 2 = 14$.If $d = -11$,$a = -11 + 3 = -8$. Then $|a| + |b| + |c| = |-8| + 4 + 2 = 14$.

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

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