Question

A plane containing the point $(3,2,0)$ and the line $\frac{x-1}{1}=\frac{y-2}{5}=\frac{z-3}{4}$ also contains the point

• A. $(0,3,1)$
• B. $(0,7,-10)$
• C. $(0,-3,1)$
• D. $(0,7,10)$

Answer 1

The correct option is D $(0,7,10)$

$\begin{array}{c}a\left(x-3\right)+b\left(y-2\right)+c\left(z-0\right)=0\\ a+5b+4c=0\\ -2a+3c=0\\ and,\\ -2a+0b+3c=0\\ \frac{a}{15-0}=\frac{-b}{3+8}=\frac{c}{0+10}\\ \frac{a}{15}=\frac{b}{-11}=\frac{c}{10}\\ 15\left(x-3\right)-11\left(y-2\right)+10z=0\\ whichclearlypassesthrough\left(0,7,10\right)\\ Hence,optionDisthecorrectanswer.\end{array}$