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# If $\overrightarrow a=(-1,1,1)\:\:\overrightarrow b=(2,0,1)\:and\:\overrightarrow x$ is vector such that $\overrightarrow x.\overrightarrow a=7,\:\overrightarrow x.\overrightarrow b=0$ and $\overrightarrow a,\overrightarrow b,\overrightarrow x$ are coplanar, then $\overrightarrow x=?$

$(a)-3\hat i+4\hat j+6\hat k\:\:\qquad\:(b)\:\:3\hat i+16\hat j-6\hat k\:\:\qquad\:(c)\:\:-3/2\hat i+5/2\hat j+3\hat k\:\:\qquad\:(d)\:\:3\hat i+4\hat j-6\hat k$

Let the vector $\overrightarrow x=x_1\hat i+x_2\hat j+x_3\hat k$
Given $\overrightarrow x,\:\overrightarrow a\:and\:\overrightarrow b$ are coplanar.
$\Rightarrow\:\overrightarrow x.(\overrightarrow a\times\overrightarrow b)=0$
$\Rightarrow\:\overrightarrow x.(\hat i+3\hat j-2\hat k)=0$
$\Rightarrow\:x_1+3x_2-2x_3=0.........(i)$
Also given that $\overrightarrow x.\overrightarrow a=7\:\:and\:\:\overrightarrow x.\overrightarrow b=0$
$\Rightarrow\:-x_1+x_2+x_3=7............(ii)$ and $2x_1+x_3=0.......(iii)$
Solving $(i)\:\:(ii)\:\:and\:\:(iii)$ we get $x_1=-\large\frac{3}{2}$$,\:x_2=\large\frac{5}{2}$$,\:x_3=3$
$\therefore\:\overrightarrow x=-\large\frac{3}{2}$$\hat i+\large\frac{5}{2}$$\hat j+3\hat k$