The equation of the plane, passing through the point $(-1,2,-3)$ and parallel to the lines. $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$, is

Question

Aicharya · Accepted Answer

Let $\left(x_1, y_1, z_1ight)=(-1,2,-3)$ $a_1, b_1, c_1=3,2,-4$ and $a_2, b_2, c_2=2,-3,2$ $	herefore \quad$ The equation of required plane is $\begin{aligned} & \left|\begin{array}{ccc}x-x_1 & y-y_1 & z-z_1 \ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}ight|=0 \ & \Rightarrow\left|\begin{array}{ccc}x+1 & y-2 & \mathrm{z}+3 \ 3 & 2 & -4 \ 2 & -3 & 2\end{array}ight|=0\end{aligned}$ $\begin{aligned} & \Rightarrow(x+1)(-8)-(y-2)(14)+(z+3)(-13)=0 \ & \Rightarrow-8 x-8-14 y+28-13 z-39=0 \ & \Rightarrow-8 x-14 y-13 z-19=0 \ & \Rightarrow 8 x+14 y+13 z+19=0\end{aligned}$

The equation of the plane, passing through the point $(-1,2,-3)$ and parallel to the lines.…

Solution