If vec{r} = 3hat{i} + 2hat{j} - 5hat{k},vec{a | vedclass

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(A) Given $\vec{r} = \lambda\vec{a} + \mu\vec{b} + \gamma\vec{c}$.Substituting the vectors: $3\hat{i} + 2\hat{j} - 5\hat{k} = \lambda(2\hat{i} - \hat{

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$\mu, \frac{\lambda}{2}, \gamma$ are in $A.P.$

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(A) Given $\vec{r} = \lambda\vec{a} + \mu\vec{b} + \gamma\vec{c}$.Substituting the vectors: $3\hat{i} + 2\hat{j} - 5\hat{k} = \lambda(2\hat{i} - \hat{j} + \hat{k}) + \mu(\hat{i} + 3\hat{j} - 2\hat{k}) + \gamma(-2\hat{i} + \hat{j} - 3\hat{k})$.Equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$:$2\lambda + \mu - 2\gamma = 3$  $(1)$$-\lambda + 3\mu + \gamma = 2$  $(2)$$\lambda - 2\mu - 3\gamma = -5$  $(3)$Solving these equations:From $(2)$,$\gamma = 2 + \lambda - 3\mu$.Substitute $\gamma$ into $(1)$ and $(3)$:$2\lambda + \mu - 2(2 + \lambda - 3\mu) = 3 \implies 7\mu = 7 \implies \mu = 1$.Substitute $\mu = 1$ into $(2)$ and $(3)$:$-\lambda + 3 + \gamma = 2 \implies \gamma - \lambda = -1 \implies \gamma = \lambda - 1$.$\lambda - 2 - 3(\lambda - 1) = -5 \implies -2\lambda + 1 = -5 \implies -2\lambda = -6 \implies \lambda = 3$.Then $\gamma = 3 - 1 = 2$.Thus,$\lambda = 3, \mu = 1, \gamma = 2$.Checking the options: $2\mu = 2(1) = 2$. The sequence $2\mu, \lambda, \gamma$ is $2, 3, 2$ (not $A$.$P$.).Checking $\mu, \frac{\lambda}{2}, \gamma$: $1, 1.5, 2$. This is an $A.P.$ with common difference $0.5$.

If $\vec{r} = 3\hat{i} + 2\hat{j} - 5\hat{k}$,$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$,and $\vec{c} = -2\hat{i} + \hat{j} - 3\hat{k}$ such that $\vec{r} = \lambda\vec{a} + \mu\vec{b} + \gamma\vec{c}$,then -

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