# If the equation of the plane passing

Question:

If the equation of the plane passing through the line of intersection of the planes

$2 x-7 y+4 z-3=0,3 x-5 y+4 z+11=0$

and the point $(-2,1,3)$ is $a x+b y+c z-7=0$, then the value of $2 \mathrm{a}+\mathrm{b}+\mathrm{c}-7$ is

Solution:

Required plane is $\mathrm{p}_{1}+\lambda \mathrm{p}_{2}=(2+3 \lambda) \mathrm{x}-(7+5 \lambda) \mathrm{y}$

$+(4+4 \lambda) \mathrm{z}-3+11 \lambda=0$

which is satisfied by $(-2,1,3)$.

Hence, $\lambda=\frac{1}{6}$

Thus, plane is $15 x-47 y+28 z-7=0$

So, $2 a+b+c-7=4$