Let the position vectors of the three points be, $\vec{a}$ = $\hat{i}+\hat{j}-\hat{2k}$ , $\vec{b}$ = $\hat{2i}-\hat{j}+\hat{k}$ and $\vec{c}$ = $\hat{i}+\hat{2j}+\hat{k}$. So, the equation of the plane passing through the points $\vec{a},\vec{b}$ and $\vec{c}$ is $(\vec{r}-\vec{a})\cdot[(\vec{b}-\vec{c})\times(\vec{c}-\vec{a})]$ = 0 ⇒ $[\vec{r}-\hat{i}+j\hat{\rightarrow}+\hat{2k}]$ . $[\hat{i}-\hat{3j}\times\hat{j}+\hat{3k}]$ = 0 ⇒ $[\vec{r}-\hat{i}+j\hat{\rightarrow}+\hat{2k}]$ . $\hat{k}-\hat{3j}-\hat{9i}$ = 0 ⇒ $\vec{r}\cdot(\hat{9i}+\hat{3j}-\hat{k})$ = 14 ... (1) So, the vector equation of the required plane is $\vec{r}\cdot(\hat{9i}+\hat{3j}-\hat{k})$ = 14 The equation of the given line is $\vec{r}$ = $\hat{3i}-\hat{j}-\hat{k}+\lambda(\hat{2i}-\hat{2j}+\hat{k})$ Position vector of any point on the given line is $\vec{r}$ = 3 + 2λ $\hat{i}$ + (- 1 - 2λ) $\hat{j}$ + (- 1 + λ) $\hat{k}$ ... (2) The point (2) lies on plane (1) if, = 14 ⇒ 9 (3) + 2λ + 3 - 1 - 2λ - (-1) + λ = 14 ⇒ 11λ + 25 = 14 ⇒ λ = - 1 Putting λ = - 1 in (2), we have $\vec{r}$ = (3 + 2λ) $\hat{i}$ + (- 1 - 2λ) $\hat{j}$ + (- 1 + λ) $\hat{k}$ = (3 + 2 - 1) $\hat{i}$ + (- 1 - 2 - 1) $\hat{j}$ + (- 1 + (- 1)) $\hat{k}$ = $\hat{i}+\hat{j}-\hat{2k}$ Thus, the position vector of the point of intersection of the given line and plane (1) is $\hat{i}+\hat{j}-\hat{2k}$ and its co-ordinates are 1, 1, - 2 .