Question

The point of intersection of the line $\frac{x-1}{3}=\frac{y+2}{4}=\frac{x-3}{-2}$ and plane $2x-y+3z=0$ is

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

Answer 1

The correct option is A $(10,-10,3)$

$\frac{x-1}{3}=\frac{y+2}{4}=\frac{x-3}{-2}=r$

$(3r+1,4r-2,-2r+3)$

$2x-y+3z=0$

$6r+24r+2-6r+9=0$

$-4r+13=0r=13/4$

$(10,-10,3)$