A) 12
B) -8
C) -4
D) 4
Correct Answer: C
Solution :
\[\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\ {{z}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{x}_{3}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\ \end{matrix} \right|=0\] \[\left| \begin{matrix} x+2 & y+2 & z-2 \\ 1+2 & -1+2 & 2-2 \\ 1+2 & 1+2 & 1-2 \\ \end{matrix} \right|=0\] \[\left| \begin{matrix} x+2 & y+2 & z+2 \\ 3 & 1 & 0 \\ 3 & 3 & -1 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[(z-2)(9-3)-1(x+2-3y-6)=0\] \[\Rightarrow \]\[6z-12-x-2+3y+6=0\] \[\Rightarrow \]\[-x+3y+6z-8=0\] \[\Rightarrow \]\[\frac{x}{8}-\frac{3y}{8}-\frac{6z}{8}+\frac{8}{8}=0\] \[\Rightarrow \]\[\frac{x}{8}-\frac{y}{\frac{8}{3}}-\frac{z}{\frac{8}{6}}=1\] \[\Rightarrow \]\[\frac{x}{-8}+\frac{9}{\frac{8}{3}}+\frac{z}{\frac{8}{6}}=1\] Sum of intercepts \[=-8+\frac{8}{3}+\frac{8}{6}\] \[=-8+\frac{16+8}{6}\] \[=-8+\frac{24}{6}\] \[=-8+4\] \[=-4\]You need to login to perform this action.
You will be redirected in
3 sec