Let $\overrightarrow a = \overrightarrow j - \overrightarrow k $ and $\overrightarrow c = \overrightarrow i - \overrightarrow j - \overrightarrow k $. Then vector $\overrightarrow b $ satisfying $(\overrightarrow a \times \overrightarrow b ) + \overrightarrow c = \overrightarrow 0 $and $\overrightarrow a .\overrightarrow b = 3$ is:
$
  A)\,2\overrightarrow i - \overrightarrow j + 2\overrightarrow k \\
  B)\,\overrightarrow i - \overrightarrow j - 2\overrightarrow k \\
  C)\,\overrightarrow i + \overrightarrow j - 2\overrightarrow k \\
  D)\, - \overrightarrow i + \overrightarrow j - 2\overrightarrow k \\
 $

Hint:In order to solve this question, assume $\overrightarrow b $ to be a variable vector. Then use $\overrightarrow a .\overrightarrow b = 3$to get an equation among the variables. Next use $(\overrightarrow a \times \overrightarrow b ) + \overrightarrow c = \overrightarrow 0 $ and get the other equation. Solve the two equations and get your answer.

Complete step-by-step answer:
Let us assume that $\overrightarrow b = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k $
As given in the question, $\overrightarrow a = \overrightarrow j - \overrightarrow k $
Now, $\overrightarrow a .\overrightarrow b = 3$
We know that dot-product is the multiplication of vectors in the same direction.
Therefore,
\[
   \Rightarrow \left( {\overrightarrow j - \overrightarrow k } \right).\left( {x\overrightarrow i + y\overrightarrow j + z\overrightarrow k } \right) = 3 \\
   \Rightarrow 0.x + 1.y - 1.z = 3 \\
   \Rightarrow y - z = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to (1) \\
 \]
We have got one equation in \[y\,\&\,z\]. Now, we will use $(\overrightarrow a \times \overrightarrow b ) + \overrightarrow c = \overrightarrow 0 $ to obtain the second equation
Let us first find $(\overrightarrow a \times \overrightarrow b )$
We know that cross-product is the multiplication of vectors in different directions.
Therefore,
\[
  (\overrightarrow a \times \overrightarrow b ) = \left( {\overrightarrow j - \overrightarrow k } \right) \times \left( {x\overrightarrow i + y\overrightarrow j + z\overrightarrow k } \right) \\
   = \left( {\begin{array}{*{20}{c}}
  {\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k } \\
  0&1&{ - 1} \\
  x&y&z
\end{array}} \right) \\
   = \overrightarrow i (z + y) - \overrightarrow j (0 + x) + \overrightarrow k (0 - x) \\
   = \overrightarrow i (z + y) - \overrightarrow j (x) - \overrightarrow k (x) \\
 \]
As given in the question , $\overrightarrow c = \overrightarrow i - \overrightarrow j - \overrightarrow k $
Therefore, using the formula $(\overrightarrow a \times \overrightarrow b ) + \overrightarrow c = \overrightarrow 0 $ we get write the above equations as
\[
   \Rightarrow \overrightarrow i (z + y) - \overrightarrow j (x) - \overrightarrow k (x) + \overrightarrow i - \overrightarrow j - \overrightarrow k = 0\overrightarrow i + 0\overrightarrow j + 0\overrightarrow k \\
   \Rightarrow \overrightarrow i (z + y + 1) - \overrightarrow j (x + 1) - \overrightarrow k (x + 1) = 0\overrightarrow i + 0\overrightarrow j + 0\overrightarrow k \]
Comparing the coefficients from both sides
$
   \Rightarrow y + z + 1 = 0\,\,,\,\,x + 1 = 0\,\,,\,\,x = -1 \\
   \Rightarrow y + z = - 1 - - - - - - - - - - - - - (2)$

Adding equations (1) and (2) we get
$
  2y = 2 \\
  y = 1
 $
Using this value of $y = 1$ and substituting in (2) we get,
$z = - 2$
Substituting value of $x$, $y$ and $z$ i.e $x= -1$ , $y= 1$ and $z= -2$ in $\overrightarrow b = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k $

We get ,$\overrightarrow b = - \overrightarrow i + \overrightarrow j - 2\overrightarrow k $

So, the correct answer is “Option D”.

Note:Another method to solve this question would be taking cross product of$\overrightarrow a $ on both sides of equation $(\overrightarrow a \times \overrightarrow b ) + \overrightarrow c = \overrightarrow 0 $which will make $\overrightarrow a \times (\overrightarrow a \times \overrightarrow b ) + \overrightarrow a \times \overrightarrow c = \overrightarrow 0 $
Then use the formula for $\overrightarrow a \times (\overrightarrow a \times \overrightarrow b )$ which is $\overrightarrow a \times (\overrightarrow a \times \overrightarrow b ) = (\overrightarrow a .\overrightarrow b )\overrightarrow a - (\overrightarrow a .\overrightarrow a )\overrightarrow b $and then solve the question. This is a much shorter method but a bit complex so you need to practice it a lot before using it in questions.