If the equation of the plane passing through the line of intersection of the planes 2x-y+z…

Question

If the equation of the plane passing through the line of intersection of the planes $2x-y+z=3$, $4x-3y+5z+9=0$, and parallel to the line $\dfrac{x+1}{-2}=\dfrac{y+3}{4}=\dfrac{z-2}{5}$, is $ax+by+cz+6=0$, then $a+b+c$ is equal to:

Accepted Answer

Family: $(2+4\lambda)x-(1+3\lambda)y+(1+5\lambda)z=(3-9\lambda)$. Perpendicular to the plane's normal is parallel to the given line's direction $(-2,4,5)$:
$$-2(2+4\lambda)-4(1+3\lambda)+5(1+5\lambda)=0 \implies 5\lambda-3=0 \implies \lambda=\frac35.$$
$$P:\ 22x-14y+20z=-12 \implies 11x-7y+10z+6=0.$$
$$a=11,\ b=-7,\ c=10 \implies a+b+c=14.$$

Solutions · 1