Let the points be $\mathbf{A}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{B}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and plane through points are $\mathbf{P}_1=\hat{\mathbf{i}}, \mathbf{P}_2=2 \hat{\mathbf{j}}, \mathbf{P}_3=3 \hat{\mathbf{k}}$

Let the point on the line be

$$ \begin{aligned} & \mathbf{r}(t)=\mathbf{A}+t(\mathbf{B}-\mathbf{A}) \\ & \Rightarrow \mathbf{r}(t)=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+t[(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & \quad \quad(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})] \\ & \quad=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+t(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & \quad=(1+t) \hat{\mathbf{i}}+(2-3) \hat{\mathbf{j}}+(1-2 t) \hat{\mathbf{k}} \end{aligned} $$

Now, vectors in the plane are

$$ \begin{aligned} & \mathbf{v}_1=\mathbf{P}_2-\mathbf{P}_1=2 \hat{\mathbf{j}}-\hat{\mathbf{i}} \\ & \mathbf{v}_2=\mathbf{P}_3-\mathbf{P}_1=3 \hat{\mathbf{k}}-\hat{\mathbf{i}} \end{aligned} $$

Normal vectors, $\mathbf{n}=\mathbf{v}_1 \times \mathbf{v}_2$

$$ \begin{aligned} & \begin{array}{r} \Rightarrow\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 2 & 0 \\ -1 & 0 & 3 \end{array}\right|=(6-0) \hat{\mathbf{i}}-(-3-0) \hat{\mathbf{j}} +(0+2 \hat{\mathbf{k}} \end{array} \\ & \Rightarrow \quad 6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $$

Using point $\mathbf{P}_1=\hat{\mathbf{i}}$ to get plane equation is

$$ \mathbf{n} \cdot(\mathbf{r}-\hat{\mathbf{i}})=0 $$

$$ \begin{array}{ll} \Rightarrow & \langle 6,3,2\rangle \cdot\langle x-1, y, z\rangle=0 \\ \Rightarrow & 6 x-6+3 y+2 z=0 \\ \Rightarrow & 6 x+3 y+2 z=6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $$

Put the value of $\mathbf{r}(t)=(1+t, 2-3 t, 1-2 t)$ into the plane Eq. (i), we get

$$ \begin{aligned} & 6 x+3 y+2 z=6 \\ & \Rightarrow 6(1+t)+3(2-3 t)+2(1-2 t)=6 \\ & \Rightarrow 6+6 t+6-9 t+2-4 t=6 \\ & \Rightarrow-7 t+14=6 \Rightarrow 7 t=8 \\ & \Rightarrow \quad t=\frac{8}{7} \\ & \text { So, } \mathbf{r}\left(\frac{8}{7}\right)=\left(1+\frac{8}{7}\right) \hat{\mathbf{i}}+\left(2-3 \times \frac{8}{7}\right) \hat{\mathbf{j}} +\left(1-2 \times \frac{8}{7}\right) \hat{\mathbf{k}} \\ & \Rightarrow \frac{15}{7} \hat{\mathbf{i}}-\frac{10}{7} \hat{\mathbf{j}}-\frac{9}{7} \hat{\mathbf{k}} \end{aligned} $$

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