If the plane passing through the points $\overline{\mathrm{i} }+\overline{ \mathrm{j}} +\overline{ \mathrm{k}}, 2\overline{\mathrm{i}} -\overline{ \mathrm{k}}$ and the origin meets the line passing through the points $\overline{\mathrm{i}}+3\overline{\mathrm{j}}-2\overline{\mathrm{k}}$ and $\overline{\mathrm{i}}-\overline{\mathrm{j}}+3\overline{\mathrm{k}}$ at the point A, then A =

$\text{ Equation of plane passing through points }(\hat{i}+\hat{j}+\hat{k}),(2\hat{i}-\hat{k})\text{ and origin is }$

$\left|\begin{array}{ccc}x& y& z\\ 1& 1& 1\\ 2& 0& -1\end{array}\right|=0$

$\Rightarrow -x+y(2+1)+z(-2)=0$

$\Rightarrow x-3y+2z=0 ...\left(i\right)$

Eq. of the line passing th

$$\begin{gathered} (\hat{i}+3\hat{j}-2\hat{k})\text{ and }(\hat{i}-\hat{j}+3\hat{k})\text{ is } \\ \frac{x-1}{0}=\frac{y-3}{4}=\frac{z+2}{-5}=\lambda \left(\text{ Let }\right) \end{gathered}$$

$\text{ So, point }P(l,4\lambda +3,-5\lambda -2)\text{ lie on the line (ii)}$

$1-12 \lambda -9-10 \lambda -4=0$

$$\begin{gathered} -22\lambda =12 \\ \lambda =\frac{-6}{11} \end{gathered}$$

So, point  $P\left(1,\frac{-24+33}{11},\frac{30-22}{11}\right)=P\left(1,\frac{9}{11},\frac{8}{11}\right)$
So, the vector form of point is  $\frac{1}{11}(11\hat{ı}+9\hat{ȷ}+8\hat{k})$.