Question

Assertion (A):

The equation of the plane through $x+y+z=6$ and $2x+3y+4z+5=0$ and passing through the point $(4,4,4)$ is $29x+23y+17z=276$.
Reason (R):

Equation of the plane through the line of intersection of the planes $P_1=0$ and $P_2=0$ is $P_1+\lambda P_2=0(\lambda \ne 0)$.

• A. Both A and R are individually true and R is the correct explanation of A.
• B. Both A and R individually true but R is not the explanation of A.
• C. A is true but R is false.
• D. A is false but R is true.

Answer 1

Answer: (A)

Both A and R are individually true and R is the correct explanation of A.

Equation of the plane through the line of intersection of the planes $P_1=0$ and $P_2=0$ is $P_1+\lambda P_2=0(\lambda \ne 0)$
However, there are infinite such planes existing for each value of $\lambda$.

Hence, to satisfy the condition of it passing through $(4,4,4)$, there has to be a particular value of $\lambda$.
$(1+2\lambda )x+(1+3\lambda )y+(1+4\lambda )z-6+5\lambda =0$

Putting $x=4,y=4,z=4$, we have
$(1+2\lambda )4+(1+3\lambda )4+(1+4\lambda )4-6+5\lambda =0$

This gives,

$\lambda =-\frac{6}{41}$
$(1+2(-\frac{6}{41}))x+(1+3(-\frac{6}{41}))y+(1+4(-\frac{6}{41}))z-6+5(-\frac{6}{41})=0$

$\frac{29x}{41}+\frac{23y}{41}+\frac{17z}{41}-\frac{276}{41}=0$

Hence the plane is,
$29x+23y+17z=276$

Hence, option A.