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(C) Given vectors are $\vec{a}=2 \hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=-2 \hat{i}+5 \hat{j}-3 \hat{k}$.Sum of the vectors is $\vec{s} = \vec{a} +

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Parallel to $YZ$-plane

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(C) Given vectors are $\vec{a}=2 \hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=-2 \hat{i}+5 \hat{j}-3 \hat{k}$.Sum of the vectors is $\vec{s} = \vec{a} + \vec{b} = (2-2) \hat{i} + (-2+5) \hat{j} + (5-3) \hat{k} = 0 \hat{i} + 3 \hat{j} + 2 \hat{k} = 3 \hat{j} + 2 \hat{k}$.Since the $\hat{i}$ component of the resulting vector $\vec{s}$ is $0$, the vector lies in the $YZ$-plane.$A$ vector that lies in a plane is parallel to that plane.Therefore, the vector is parallel to the $YZ$-plane.

The vector in the direction of the sum of the vectors $\vec{a}=2 \hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=-2 \hat{i}+5 \hat{j}-3 \hat{k}$ is

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