\displaystyle \textbf{Question 1: }~\text{Find the vector equation of the following planes in scalar product form }(\overrightarrow{r}\cdot\overrightarrow{n}=a):
\displaystyle \text{(i) }\ \overrightarrow{r}=(2\widehat{i}-\widehat{k})+\lambda\widehat{i}+\mu(\widehat{i}-2\widehat{j}-\widehat{k})
\displaystyle \text{(ii) }\ \overrightarrow{r}=(1+s-t)\widehat{i}+(2-s)\widehat{j}+(3-2s+2t)\widehat{k}
\displaystyle \text{(iii) }\ \overrightarrow{r}=(\widehat{i}+\widehat{j})+\lambda(\widehat{i}+2\widehat{j}-\widehat{k})+\mu(-\widehat{i}+\widehat{j}-2\widehat{k})
\displaystyle \text{(iv) }\ \overrightarrow{r}=\widehat{i}-\widehat{j}+\lambda(\widehat{i}+\widehat{j}+\widehat{k})+\mu(4\widehat{i}-2\widehat{j}+3\widehat{k})
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{We know that the equation } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}+\mu\overrightarrow{c} \text{ represents a plane passing through a} \\ \text{point whose position vector is } \overrightarrow{a} \text{ and parallel to the vectors } \overrightarrow{b} \text{ and } \overrightarrow{c}.
\displaystyle \text{Here, } \overrightarrow{a}=2\widehat{i}+0\widehat{j}-\widehat{k},\ \overrightarrow{b}=\widehat{i},\ \overrightarrow{c}=\widehat{i}-2\widehat{j}-\widehat{k}.
\displaystyle \text{Normal vector, } \overrightarrow{n}=\overrightarrow{b}\times\overrightarrow{c}.
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&0&0\\1&-2&-1\end{vmatrix}
\displaystyle =\widehat{i}\big(0(-1)-0(-2)\big)-\widehat{j}\big(1(-1)-0(1)\big)+\widehat{k}\big(1(-2)-0(1)\big)
\displaystyle =0\widehat{i}+\widehat{j}-2\widehat{k}.
\displaystyle =\widehat{j}-2\widehat{k}.
\displaystyle \text{The vector equation of the plane in scalar product form is}
\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(\widehat{j}-2\widehat{k})=(2\widehat{i}+0\widehat{j}-\widehat{k})\cdot(\widehat{j}-2\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot(\widehat{j}-2\widehat{k})=2.

\displaystyle \text{(ii)  }
\displaystyle \text{The given equation of the plane is}
\displaystyle \overrightarrow{r}=(1+s-t)\widehat{i}+(2-s)\widehat{j}+(3-2s+2t)\widehat{k}.
\displaystyle \Rightarrow \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+s(\widehat{i}-\widehat{j}-2\widehat{k})+t(-\widehat{i}+0\widehat{j}+2\widehat{k}).
\displaystyle \text{We know that the equation } \overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b}+t\overrightarrow{c} \text{ represents a plane passing through a} \\ \text{point whose position vector is } \overrightarrow{a} \text{ and parallel to the vectors } \overrightarrow{b} \text{ and } \overrightarrow{c}.
\displaystyle \text{Here, } \overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{b}=\widehat{i}-\widehat{j}-2\widehat{k},\ \overrightarrow{c}=-\widehat{i}+0\widehat{j}+2\widehat{k}.
\displaystyle \text{Normal vector, } \overrightarrow{n}=\overrightarrow{b}\times\overrightarrow{c}.
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-1&-2\\-1&0&2\end{vmatrix}
\displaystyle =\widehat{i}\big((-1)(2)-(-2)(0)\big)-\widehat{j}\big(1(2)-(-2)(-1)\big)+\widehat{k}\big(1(0)-(-1)(-1)\big)
\displaystyle =-2\widehat{i}+0\widehat{j}-\widehat{k}.
\displaystyle =-2\widehat{i}-\widehat{k}.
\displaystyle \text{The vector equation of the plane in scalar product form is}
\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(-2\widehat{i}-\widehat{k})=(\widehat{i}+2\widehat{j}+3\widehat{k})\cdot(-2\widehat{i}-\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(-1(2\widehat{i}+\widehat{k})\big)=-2+0-3.
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(-1(2\widehat{i}+\widehat{k})\big)=-5.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(2\widehat{i}+\widehat{k})=5.

\displaystyle \text{(iii)  }
\displaystyle \text{We know that the equation } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}+\mu\overrightarrow{c} \text{ represents a plane passing through a} \\ \text{point whose position vector is } \overrightarrow{a} \text{ and parallel to the vectors } \overrightarrow{b} \text{ and } \overrightarrow{c}.
\displaystyle \text{Here, } \overrightarrow{a}=\widehat{i}+\widehat{j}+0\widehat{k},\ \overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k},\ \overrightarrow{c}=-\widehat{i}+\widehat{j}-2\widehat{k}.
\displaystyle \text{Normal vector, } \overrightarrow{n}=\overrightarrow{b}\times\overrightarrow{c}.
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&-1\\-1&1&-2\end{vmatrix}
\displaystyle =\widehat{i}\big(2(-2)-(-1)(1)\big)-\widehat{j}\big(1(-2)-(-1)(-1)\big)+\widehat{k}\big(1(1)-2(-1)\big)
\displaystyle =-3\widehat{i}+3\widehat{j}+3\widehat{k}.
\displaystyle \text{The vector equation of the plane in scalar product form is}
\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(-3\widehat{i}+3\widehat{j}+3\widehat{k})=(\widehat{i}+\widehat{j}+0\widehat{k})\cdot(-3\widehat{i}+3\widehat{j}+3\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot(-3\widehat{i}+3\widehat{j}+3\widehat{k})=-3+3.
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(3(-\widehat{i}+\widehat{j}+\widehat{k})\big)=0.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(-\widehat{i}+\widehat{j}+\widehat{k})=0.

\displaystyle \text{(iv)  }
\displaystyle \text{We know that the equation } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}+\mu\overrightarrow{c} \text{ represents a plane passing through a} \\ \text{point whose position vector is } \overrightarrow{a} \text{ and parallel to the vectors } \overrightarrow{b} \text{ and } \overrightarrow{c}.
\displaystyle \text{Here, } \overrightarrow{a}=\widehat{i}-\widehat{j}+0\widehat{k},\ \overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{c}=4\widehat{i}-2\widehat{j}+3\widehat{k}.
\displaystyle \text{Normal vector, } \overrightarrow{n}=\overrightarrow{b}\times\overrightarrow{c}.
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&1&1\\4&-2&3\end{vmatrix}
\displaystyle =\widehat{i}\big(1\cdot3-1(-2)\big)-\widehat{j}\big(1\cdot3-1\cdot4\big)+\widehat{k}\big(1(-2)-1\cdot4\big)
\displaystyle =5\widehat{i}+\widehat{j}-6\widehat{k}.
\displaystyle \text{The vector equation of the plane in scalar product form is}
\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(5\widehat{i}+\widehat{j}-6\widehat{k})=(\widehat{i}-\widehat{j}+0\widehat{k})\cdot(5\widehat{i}+\widehat{j}-6\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot(5\widehat{i}+\widehat{j}-6\widehat{k})=5-1+0.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(5\widehat{i}+\widehat{j}-6\widehat{k})=4.

\displaystyle \textbf{Question 2: }~\text{Find the cartesian form of the equation of the following planes:}
\displaystyle \text{(i) }\ \overrightarrow{r}=(\widehat{i}-\widehat{j})+s(-\widehat{i}+\widehat{j}+2\widehat{k})+t(\widehat{i}+2\widehat{j}+\widehat{k})
\displaystyle \text{(ii) }\ \overrightarrow{r}=(1+s+t)\widehat{i}+(2-s+t)\widehat{j}+(3-2s+2t)\widehat{k}
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \overrightarrow{r}=(\widehat{i}-\widehat{j}+0\widehat{k})+s(-\widehat{i}+\widehat{j}+2\widehat{k})+t(\widehat{i}+2\widehat{j}+\widehat{k}).
\displaystyle \text{We know that the equation } \overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b}+t\overrightarrow{c} \text{ represents a plane passing through a point} \\ \text{whose position vector is } \overrightarrow{a} \text{ and parallel to the vectors } \overrightarrow{b} \text{ and } \overrightarrow{c}.
\displaystyle \text{Here, } \overrightarrow{a}=\widehat{i}-\widehat{j}+0\widehat{k},\ \overrightarrow{b}=-\widehat{i}+\widehat{j}+2\widehat{k},\ \overrightarrow{c}=\widehat{i}+2\widehat{j}+\widehat{k}.
\displaystyle \text{Normal vector, } \overrightarrow{n}=\overrightarrow{b}\times\overrightarrow{c}.
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-1&1&2\\1&2&1\end{vmatrix}
\displaystyle =\widehat{i}\big(1\cdot1-2\cdot2\big)-\widehat{j}\big((-1)(1)-2(1)\big)+\widehat{k}\big((-1)(2)-1(1)\big)
\displaystyle =-3\widehat{i}+3\widehat{j}-3\widehat{k}.
\displaystyle \text{The vector equation of the plane in scalar product form is}
\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(-3\widehat{i}+3\widehat{j}-3\widehat{k})=(\widehat{i}-\widehat{j}+0\widehat{k})\cdot(-3\widehat{i}+3\widehat{j}-3\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(-3(\widehat{i}-\widehat{j}+\widehat{k})\big)=-3-3+0.
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(-3(\widehat{i}-\widehat{j}+\widehat{k})\big)=-6.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+\widehat{k})=2.
\displaystyle \text{For Cartesian form, substitute } \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}. \text{ Then, we get}
\displaystyle (x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(\widehat{i}-\widehat{j}+\widehat{k})=2.
\displaystyle \Rightarrow x-y+z=2.

\displaystyle \text{(ii)  }
\displaystyle \text{The given equation of the plane is}
\displaystyle \overrightarrow{r}=(1+s+t)\widehat{i}+(2-s+t)\widehat{j}+(3-2s+2t)\widehat{k}.
\displaystyle \Rightarrow \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+s(\widehat{i}-\widehat{j}-2\widehat{k})+t(\widehat{i}+\widehat{j}+2\widehat{k}).
\displaystyle \text{We know that the equation } \overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b}+t\overrightarrow{c} \text{ represents a plane passing through a point} \\ \text{whose position vector is } \overrightarrow{a} \text{ and parallel to the vectors } \overrightarrow{b} \text{ and } \overrightarrow{c}.
\displaystyle \text{Here, } \overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{b}=\widehat{i}-\widehat{j}-2\widehat{k},\ \overrightarrow{c}=\widehat{i}+\widehat{j}+2\widehat{k}.
\displaystyle \text{Normal vector, } \overrightarrow{n}=\overrightarrow{b}\times\overrightarrow{c}.
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-1&-2\\1&1&2\end{vmatrix}
\displaystyle =\widehat{i}\big((-1)(2)-(-2)(1)\big)-\widehat{j}\big(1(2)-(-2)(1)\big)+\widehat{k}\big(1(1)-(-1)(1)\big)
\displaystyle =0\widehat{i}-4\widehat{j}+2\widehat{k}.
\displaystyle =-4\widehat{j}+2\widehat{k}.
\displaystyle \text{The vector equation of the plane in scalar product form is}
\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(-4\widehat{j}+2\widehat{k})=(\widehat{i}+2\widehat{j}+3\widehat{k})\cdot(-4\widehat{j}+2\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(-2(2\widehat{j}-\widehat{k})\big)=0-8+6.
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(-2(2\widehat{j}-\widehat{k})\big)=-2.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(2\widehat{j}-\widehat{k})=1.
\displaystyle \text{For Cartesian form, substitute } \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}. \text{ Then, we get}
\displaystyle (x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(2\widehat{j}-\widehat{k})=1.
\displaystyle \Rightarrow 2y-z=1.

\displaystyle \textbf{Question 3: }~\text{Find the vector equation of the following planes in non-parametric form:}
\displaystyle \text{(i) }\ \overrightarrow{r}=(\lambda-2\mu)\widehat{i}+(3-\mu)\widehat{j}+(2\lambda+\mu)\widehat{k}
\displaystyle \text{(ii) }\ \overrightarrow{r}=(2\widehat{i}+2\widehat{j}-\widehat{k})+\lambda(\widehat{i}+2\widehat{j}+3\widehat{k})+\mu(5\widehat{i}-2\widehat{j}+7\widehat{k})
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{The given equation of the plane is}
\displaystyle \overrightarrow{r}=(1+s+t)\widehat{i}+(2-s+t)\widehat{j}+(3-2s+2t)\widehat{k}.
\displaystyle \Rightarrow \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+s(\widehat{i}-\widehat{j}-2\widehat{k})+t(\widehat{i}+\widehat{j}+2\widehat{k}).
\displaystyle \text{We know that the equation } \overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b}+t\overrightarrow{c} \text{ represents a plane passing} \\ \text{through a point whose position vector is } \overrightarrow{a} \text{ and parallel to the vectors } \overrightarrow{b} \text{ and } \overrightarrow{c}.
\displaystyle \text{Here, } \overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{b}=\widehat{i}-\widehat{j}-2\widehat{k},\ \overrightarrow{c}=\widehat{i}+\widehat{j}+2\widehat{k}.
\displaystyle \text{Normal vector, } \overrightarrow{n}=\overrightarrow{b}\times\overrightarrow{c}.
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-1&-2\\1&1&2\end{vmatrix}
\displaystyle =\widehat{i}\big((-1)(2)-(-2)(1)\big)-\widehat{j}\big(1(2)-(-2)(1)\big)+\widehat{k}\big(1(1)-(-1)(1)\big)
\displaystyle =0\widehat{i}-4\widehat{j}+2\widehat{k}.
\displaystyle =-4\widehat{j}+2\widehat{k}.
\displaystyle \text{The vector equation of the plane in scalar product form is}
\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(-4\widehat{j}+2\widehat{k})=(\widehat{i}+2\widehat{j}+3\widehat{k})\cdot(-4\widehat{j}+2\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(-2(2\widehat{j}-\widehat{k})\big)=0-8+6.
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(-2(2\widehat{j}-\widehat{k})\big)=-2.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(2\widehat{j}-\widehat{k})=1.
\displaystyle \text{For Cartesian form, substitute } \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}. \text{ Then, we get}
\displaystyle (x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(2\widehat{j}-\widehat{k})=1.
\displaystyle \Rightarrow 2y-z=1.

\displaystyle \text{(ii)  }
\displaystyle \text{We know that the equation } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}+\mu\overrightarrow{c} \text{ represents a plane passing through} \\ \text{a point whose position vector is } \overrightarrow{a} \text{ and parallel to the vectors } \overrightarrow{b} \text{ and } \overrightarrow{c}.
\displaystyle \text{Here, } \overrightarrow{a}=2\widehat{i}+2\widehat{j}-\widehat{k},\ \overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{c}=5\widehat{i}-2\widehat{j}+7\widehat{k}.
\displaystyle \text{Normal vector, } \overrightarrow{n}=\overrightarrow{b}\times\overrightarrow{c}.
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&3\\5&-2&7\end{vmatrix}
\displaystyle =\widehat{i}\big(2\cdot7-3(-2)\big)-\widehat{j}\big(1\cdot7-3\cdot5\big)+\widehat{k}\big(1(-2)-2\cdot5\big)
\displaystyle =20\widehat{i}+8\widehat{j}-12\widehat{k}.
\displaystyle \text{The vector equation of the plane in scalar product form is}
\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(20\widehat{i}+8\widehat{j}-12\widehat{k})=(2\widehat{i}+2\widehat{j}-\widehat{k})\cdot(20\widehat{i}+8\widehat{j}-12\widehat{k}).
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(4(5\widehat{i}+2\widehat{j}-3\widehat{k})\big)=40+16+12.
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big(4(5\widehat{i}+2\widehat{j}-3\widehat{k})\big)=68.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(5\widehat{i}+2\widehat{j}-3\widehat{k})=17.

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