Question

If the equation of the plane passing through the line of intersection of the planes $2x–7y+4z–3=0,3x–5y+4z+11=0$and the point$(-2,1,3)$ is $ax+by+cz–7=0$, then the value of $2a+b+c–7$ is __________

Answer 1

Step 1: Solve for the equation of the plane

Let $P_1\Rightarrow 2x–7y+4z–3=0,P_2\Rightarrow 3x–5y+4z+11=0$

The equation of plane can be written using the family of planes:$P_1+\lambda P_2=0.$

$\Rightarrow 2x–7y+4z–3+\lambda \left(3x–5y+4z+11\right)=0...........\left(i\right)$

It passes through $(-2,1,3)$.

$$\begin{gathered} \Rightarrow (-4–7+12–3)+\lambda (-6–5+12+11)=0 \\ \Rightarrow –2+\lambda \left(12\right)=0 \\ \Rightarrow \lambda =\frac{1}{6} \end{gathered}$$

Now putting the value of $\lambda$ in equation $\left(i\right)$ we get

$$\begin{gathered} 2x–7y+4z–3+\lambda \left(3x–5y+4z+11\right)=0 \\ \Rightarrow 12x–42y+24z–18+3x–5y+4z+11=0 \\ \Rightarrow 15x–47y+28z–7=0 \end{gathered}$$

Therefore the equation of the plane is $15x–47y+28z–7=0$

Step 2. Find the value of given expression

Now comparing the equation $ax+by+cz–7=0$ with the equation of the plane $15x–47y+28z–7=0$

We have $a=15,b=-47$ and $c=28$

Now the value of the expression $2a+b+c–7$ is

$$\begin{gathered} 2a+b+c–7 \\ =2\times 15-47+28-7 \\ =4 \end{gathered}$$

Hence the value of $2a+b+c–7$ is $4$