IIT-JEE 2012 · Paper 2, Mathematics question 41

Family of planes through the line of intersection:
$$(x+2y+3z-2)+\lambda(x-y+z-3)=0,$$
with normal $(1+\lambda,\ 2-\lambda,\ 3+\lambda)$.

Distance from $(3,1,-1)$. The first bracket vanishes there ($3+2-3-2=0$) and the second equals $-2$:
$$\frac{2|\lambda|}{\sqrt{(1+\lambda)^2+(2-\lambda)^2+(3+\lambda)^2}}=\frac2{\sqrt3}\implies3\lambda^2=3\lambda^2+4\lambda+14\implies\lambda=-\frac72.$$

Plane. $-\frac52x+\frac{11}2y-\frac12z+\frac{17}2=0$, i.e. $5x-11y+z=17$.

(A)