JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Three Dimensional Geometry

A plane through the line of intersection of the two given planes can be written as $\mu (x+4y-z+7)+\lambda (3x+y+5z-8)=0$, whose coefficients are $(\mu +3\lambda )$ for $x$, $(4\mu +\lambda )$ for $y$, $(-\mu +5\lambda )$ for $z$ and constant $(7\mu -8\lambda )$. To match $ax+by+6z=15$, the coefficient of $z$ and the constant must be in the ratio $6:(-15)$, so $\frac{-\mu +5\lambda }{6}=\frac{7\mu -8\lambda }{-15}$, which simplifies to $\mu =-\lambda$. With $\mu =-\lambda$ the $x$-coefficient is $2\lambda$, the $y$-coefficient $-3\lambda$, the $z$-coefficient $6\lambda$ and the constant $-15\lambda$, so dividing through by $\lambda$ the plane is $2x-3y+6z=15$. The distance of $(3,2,-1)$ from it is $\frac{|2(3)-3(2)+6(-1)-15|}{\sqrt{2^2+3^2+6^2}}=\frac{|6-6-6-15|}{7}=\frac{21}{7}=3$.