Question

If a point P (3,- 4, 5) lies on the plane which passes through the intersection of two planes x-3y+4z =0 & 2x+y+6z+7=0 then the equation of this plane is -

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

We know that If two planes $a_{1}x+b_{1}y+c_{1}z+d_1=0$ & $a_{2}x+b_{2}y+c_{2}z+d_2=0$ intersect then the equation of plane passing through the line of intersection of these two planes can be written as $a_{1}x+b_{1}y+c_{1}z+d_1+\lambda (a_{2}x+b_{2}y+c_{2}z+d_2)=0$.
We’ll do that appropriate substitution -
$x-3y+4z+\lambda (2x+y+6z+7)=0$
Or $(1+2\lambda )x+(\lambda -3)y+(6\lambda +4)z+7\lambda =0$
It is given that point P (3, -4 ,5) lies on the plane. So it’ll satisfy the equation.
$(1+2\lambda )3+(\lambda -3)(-4)+(6\lambda +4)5+7\lambda =0$
$35+39\lambda =0$
Or $\lambda =\frac{-39}{35}$
So, the equation of plane will be -
$-43x-144y-94z-273=0$