Let $\overrightarrow{\mathit{c}}$ = $\mathit{x}\hat{\mathit{i}}+\mathit{y}\hat{\mathit{j}}+\mathit{z}\hat{\mathit{k}}$
$\overrightarrow{\mathit{a}}$ = $\hat{\mathit{i}}+\hat{\mathit{j}}+\hat{\mathit{k}}$
∴ $\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{c}}$ = $\left|\begin{array}{ccc}\hat{\mathit{i}}& \hat{\mathit{j}}& \hat{\mathit{k}}\\ 1& 1& 1\\ \mathit{x}& \mathit{y}& \mathit{z}\end{array}\right|$
$\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{c}}$ = $\hat{\mathit{i}}\left(\mathit{z}-\mathit{y}\right)-\hat{\mathit{j}}\left(\mathit{z}-\mathit{x}\right)+\hat{\mathit{k}}\left(\mathit{y}-\mathit{x}\right)$ ... (1)
Now, $\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{c}}$ = $\overrightarrow{\mathit{b}}$
$\overrightarrow{\mathit{b}}$ = $\hat{\mathit{j}}-\hat{\mathit{k}}$ ... (2)
Comparing (1) and (2), we get :
z – y = 0 ⇒ z = y ...(3)
z – x = -1 ...(4)
y – x = -1 ...(5)
Also, given that
$\overrightarrow{\mathit{a}}.\overrightarrow{\mathit{c}}$ = 3
∴ $\hat{\mathit{i}}+\hat{\mathit{j}}+\hat{\mathit{k}}$ . $\mathit{x}\hat{\mathit{i}}+\mathit{y}\hat{\mathit{j}}+\mathit{z}\hat{\mathit{k}}$ = 3
x + y + z = 3
Using (3), we get, x + 2y = 3 ...(6)
Adding (5) and (6), we get
3y = 2 ⇒ y = $\frac{2}{3}$
∴ z = $\frac{2}{3}$ Since z = y
From (6), we have,x = 3 - 2y
⇒ x = 3 - $\frac{2\times 2}{3}$
⇒ x = $\frac{9-4}{3}$
⇒ x = $\frac{5}{3}$
∴ $\overrightarrow{\mathit{c}}$ = $\frac{5}{3}\hat{\mathit{i}}+\frac{2}{3}\hat{\mathit{j}}+\frac{2}{3}\hat{\mathit{k}}$
Thus, the required vector $\overrightarrow{\mathit{c}}$ is $\frac{5}{3}\hat{\mathit{i}}+\frac{2}{3}\hat{\mathit{j}}+\frac{2}{3}\hat{\mathit{k}}$
OR
$\overrightarrow{\mathit{a}}+\overrightarrow{\mathit{b}}+\overrightarrow{\mathit{c}}$ = 0 ⇒ $\overrightarrow{\mathit{a}}+\overrightarrow{\mathit{b}}$ = $-\overrightarrow{\mathit{c}}$
$\overrightarrow{\mathit{a}}+\overrightarrow{\mathit{b}}.\overrightarrow{\mathit{a}}+\overrightarrow{\mathit{b}}$ = $-\overrightarrow{\mathit{c}}.-\overrightarrow{\mathit{c}}$
$\overrightarrow{\mathit{a}}.\overrightarrow{\mathit{a}}$ + $2\overrightarrow{\mathit{a}}.\overrightarrow{\mathit{b}}$ + $\overrightarrow{\mathit{b}}.\overrightarrow{\mathit{b}}$ = $\overrightarrow{\mathit{c}}.\overrightarrow{\mathit{c}}$
${\left|\overrightarrow{\mathit{a}}\right|}^2+2|\overrightarrow{\mathit{a}}||\overrightarrow{\mathit{b}}|$ cos θ + ${\left|\overrightarrow{\mathit{b}}\right|}^2$ = ${\left|\overrightarrow{\mathit{c}}\right|}^2$
$3^2+\left(2\right)\left(3\right)\left(5\right)\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }+5^2$ = $7^2$
9 + 30cos θ + 25 = 49
30 cos θ = 15 ⇒ cos θ = $\frac{1}{2}$
cos θ = cos 60° ⇒ θ = 60°
Hence proved.