$$\begin{gathered} \mathrm{If} \overrightarrow{\mathrm{a}} = \hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}} \mathrm{and} \overrightarrow{\mathrm{b}} = \hat{\mathrm{j}} - \hat{\mathrm{k}}, \mathrm{find} \mathrm{a} \mathrm{vector} \overrightarrow{\mathrm{c}} \mathrm{such} \mathrm{that} \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} = \overrightarrow{\mathrm{b}} \\ \mathrm{and} \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}} = 3 \end{gathered}$$

$$\begin{gathered} \mathrm{Let} \overrightarrow{\mathrm{c}} = \mathrm{x}\hat{\mathrm{i}} + \mathrm{y}\hat{\mathrm{j}} + \mathrm{z}\widehat{\mathrm{k} } \\ \overrightarrow{\mathrm{a}} = \hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}} \\ \therefore \overrightarrow{\mathrm{a}} \mathrm{x} \overrightarrow{\mathrm{c}} = \left[\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}}\\ 1 & 1 & 1\\ \mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right] \\ \mathrm{a} \mathrm{x} \overrightarrow{\mathrm{c}} = \hat{\mathrm{i}} ( \mathrm{z} - \mathrm{y} ) - \hat{\mathrm{j}} ( \mathrm{z} - \mathrm{x} ) + \hat{\mathrm{k}} ( \mathrm{y} - \mathrm{x} ) ...............\left(1\right) \\ \mathrm{Now}, \overrightarrow{\mathrm{a}} \mathrm{x} \overrightarrow{\mathrm{c}} = \overrightarrow{\mathrm{b}} \\ \overrightarrow{\mathrm{b}} = \hat{\mathrm{j}} - \hat{\mathrm{k}} ................\left(2\right) \end{gathered}$$

Comparing (1) and (2), we get:

$$\begin{gathered} \mathrm{z} - \mathrm{y} = 0 \Rightarrow \mathrm{z} = \mathrm{y} .............\left(3\right) \\ \mathrm{z} - \mathrm{x} = -1 .............\left(4\right) \\ \mathrm{y} - \mathrm{x} = -1 .............\left(5\right) \end{gathered}$$

Also, given that

$$\begin{gathered} \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}} =3 \\ \therefore \hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}} . \mathrm{x} \hat{\mathrm{i}} +\mathrm{y} \hat{\mathrm{j}} +\mathrm{z} \hat{\mathrm{k}} = 3 \\ \mathrm{x} + \mathrm{y} + \mathrm{z} = 3 \end{gathered}$$

Using (3), we get,  x + 2y = 3    .                  ...............(6)

Adding (5) and (6), we get

$$\begin{gathered} 4\mathrm{y} = 2 \Rightarrow \mathrm{y} = \frac{2}{3} \\ \therefore \mathrm{z} = \frac{2}{3} \because \mathrm{z}=\mathrm{y} \\ \mathrm{from} \left(6\right) \mathrm{we} \mathrm{have}, \\ \mathrm{x} = 3 - 2\mathrm{y} \\ \Rightarrow \mathrm{x} = 3 - \frac{2 \mathrm{x} 2}{3} \\ \Rightarrow \mathrm{x} = \frac{9 - 4}{3} \\ \Rightarrow \mathrm{x} = \frac{5}{3} \\ \therefore \overrightarrow{\mathrm{c}} = \frac{5}{3} \hat{\mathrm{i}} + \frac{2}{3} \hat{\mathrm{j}} + \frac{2}{3} \hat{\mathrm{k}}. \\ \mathrm{Thus} \mathrm{the} \mathrm{required} \mathrm{vector} \mathrm{is} \overrightarrow{\mathrm{c}} = \frac{5}{3} \hat{\mathrm{i}} + \frac{2}{3} \hat{\mathrm{j}} + \frac{2}{3} \hat{\mathrm{k}}. \end{gathered}$$