Let vec{a}=hat{i}+hat{j}+2hat{k},vec{b}=2hat{ | vedclass

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(D) Since $\vec{v}$ lies in the plane of $\vec{a}$ and $\vec{b}$,we can write $\vec{v} = \lambda \vec{a} + \mu \vec{b}$.Substituting the given vectors

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$9$

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(D) Since $\vec{v}$ lies in the plane of $\vec{a}$ and $\vec{b}$,we can write $\vec{v} = \lambda \vec{a} + \mu \vec{b}$.Substituting the given vectors,$\vec{v} = \lambda(\hat{i}+\hat{j}+2\hat{k}) + \mu(2\hat{i}-3\hat{j}+\hat{k}) = (\lambda+2\mu)\hat{i} + (\lambda-3\mu)\hat{j} + (2\lambda+\mu)\hat{k}$.Given $\vec{v} \cdot \hat{j} = 7$,we have $\lambda - 3\mu = 7$  (Equation $1$).The projection of $\vec{v}$ on $\vec{c}$ is $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{2}{\sqrt{3}}$.Since $|\vec{c}| = \sqrt{1^2+(-1)^2+1^2} = \sqrt{3}$,we have $\vec{v} \cdot \vec{c} = 2$.Calculating $\vec{v} \cdot \vec{c} = (\lambda+2\mu)(1) + (\lambda-3\mu)(-1) + (2\lambda+\mu)(1) = \lambda+2\mu - \lambda+3\mu + 2\lambda+\mu = 2\lambda+6\mu = 2$,so $\lambda+3\mu = 1$ (Equation $2$).Adding Equation $1$ and Equation $2$: $(\lambda-3\mu) + (\lambda+3\mu) = 7+1 \implies 2\lambda = 8 \implies \lambda = 4$.Substituting $\lambda=4$ into Equation $2$: $4+3\mu = 1 \implies 3\mu = -3 \implies \mu = -1$.Thus,$\vec{v} = 4(\hat{i}+\hat{j}+2\hat{k}) - 1(2\hat{i}-3\hat{j}+\hat{k}) = (4-2)\hat{i} + (4+3)\hat{j} + (8-1)\hat{k} = 2\hat{i}+7\hat{j}+7\hat{k}$.Finally,$\vec{v} \cdot (\hat{i}+\hat{k}) = (2\hat{i}+7\hat{j}+7\hat{k}) \cdot (\hat{i}+\hat{k}) = 2(1) + 7(0) + 7(1) = 2+7 = 9$.

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