Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}.\vec{b}=-\hat{i}+\hat{j}.$ Then a vector $\vec{c}$ satisfying $\vec{a}×\vec{c}=\vec{b}$ and $\vec{a}.\vec{c}=8$ is equal to

$3\hat{i}+3\hat{j}+2\hat{k}$

The correct answer is Option (4) → $3\hat{i}+3\hat{j}+2\hat{k}$

$\vec{a}×\vec{c}=\vec{b}$

let $\vec c=x\hat i+y\hat j+z\hat k$

$\vec a.\vec c=x+y+z=8$   ...(1)

$\vec{a}×\vec{c}=\vec{b}$

so $(z-y)\hat i+(x-z)\hat j+(y-z)\hat k\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&1\\x&y&z\end{vmatrix}=\vec b$

$=-\hat i+\hat j$

so $z=y-1$

$x-z=1$

so $x=z+1$

$y=z+1$

so $x=y$

from (1)

$y+y+y-1=8$

$y=3$

so $z=2,x=3$

$\vec c=3\hat{i}+3\hat{j}+2\hat{k}$