Let $\vec{c}$ = $x\hat{i}+y\hat{j}+z\hat{k}$ $\vec{a}$ = $\hat{i}+\hat{j}+\hat{k}$ ∴ $\vec{a}\times\vec{c}$ = $\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\x&y&z\end{vmatrix}$ $\vec{a}\times\vec{c}$ = $\hat{i}(z-y)-\hat{j}(z-x)+\hat{k}(y-x)$ ... (1) Now, $\vec{a}\times\vec{c}$ = $\vec{b}$ $\vec{b}$ = $\hat{j}-\hat{k}$ ... (2) Comparing (1) and (2), we get : z – y = 0 ⇒ z = y ...(3) z – x = -1 ...(4) y – x = -1 ...(5) Also, given that $\vec{a}\cdot\vec{c}$ = 3 ∴ $\hat{i}+\hat{j}+\hat{k}$ . $x\hat{i}+y\hat{j}+z\hat{k}$ = 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 - $\frac{2\times2}{3}$ ⇒ x = $\frac{9-4}{3}$ ⇒ x = $\frac{5}{3}$ ∴ $\vec{c}$ = $\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$ Thus, the required vector $\vec{c}$ is $\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$ OR $\vec{a}+\vec{b}+\vec{c}$ = 0 ⇒ $\vec{a}+\vec{b}$ = $-\vec{c}$ $\vec{a}+\vec{b}\cdot\vec{a}+\vec{b}$ = $-\vec{c}\cdot-\vec{c}$ $\vec{a}\cdot\vec{a}$ + $2\vec{a}\cdot\vec{b}$ + $\vec{b}\cdot\vec{b}$ = $\vec{c}\cdot\vec{c}$ $|\vec{a}|^2+2|\vec{a}||\vec{b}|$ cos θ + $|\vec{b}|^2$ = $|\vec{c}|^2$ $3^2+(2)(3)(5)\cos\theta+5^2$ = $7^2$ 9 + 30cos θ + 25 = 49 30 cos θ = 15 ⇒ cos θ = $\frac{1}{2}$ cos θ = cos 60° ⇒ θ = 60° Hence proved.