Answer :
Answer:
The normal vector is [tex]{\bf n} \ =\ 3{\bf i} -8{\bf j} +4{\bf k}[/tex]
Step-by-step explanation:
Let
[tex]{\bf b}=\langle\,x_1,\,y_1,\,z_1\, \rangle\,, \ \ {\bf r} = \langle\,x_2,\,y_2,\,z_2\, \rangle\,, \ \ {\bf s} = \langle\,x_3,\,y_3,\,z_3\, \rangle.[/tex]
The vectors
[tex]\overrightarrow{QR} \ = \ {\bf r} - {\bf b} \,, \qquad \overrightarrow{QS} \ = \ {\bf s} - {\bf b} \,,[/tex]
then lie in the plane. The normal to the plane is given by the cross product
[tex]{\bf n} = ({\bf r} - {\bf b})\times ( {\bf s} - {\bf b})[/tex]
We have the following points:
[tex]Q(-6,\,-4,\,-6)\,, \ \ R(-2,\,0,\,-1)\,, \ \ S(-2,\,1,\,1)\,.[/tex]
when the plane passes through [tex]Q,\, R[/tex], and [tex]S[/tex], then the vectors
[tex]\overrightarrow{QR} = \langle\, -2-(-6),\, 0-(-4),\, -1-(-6)\, \rangle\,, \overrightarrow{QS}= \langle\, -2-(-6),\, 1-(-4),\, 1-(-6)\, \rangle\,,\\\\\overrightarrow{QR} = \langle\, 4,\, 4,\, 5\, \rangle\,,\qquad \overrightarrow{QS}= \langle\, 4,\, 5,\, 7\, \rangle\,[/tex]
lie in the plane. Thus the cross-product
[tex]{\bf n} \ = \ \left|\begin{array}{ccc}{\bf i} & {\bf j} & {\bf k} \\ 4 & 4 & 5 \\ 4 & 5 & 7 \end{array}\right| \ = \begin{pmatrix}4\cdot \:7-5\cdot \:5&5\cdot \:4-4\cdot \:7&4\cdot \:5-4\cdot \:4\end{pmatrix} \ = \ 3{\bf i} -8{\bf j} +4{\bf k}[/tex]
is normal to the plane.