Manipal Entrance Test 2015 Math Paper

Let $P(3,8,2)$ be the given point. Let the equation of a line through $P(3,8,2)$ and parallel to the plane $3x+2y-2z+15=0$ be $\frac{x-3}{a}=\frac{y-8}{b}=\frac{z-2}{c}$ ....(i) Then, normal to the plane is perpendicular to the parallel line. Then, $3a+2b-2c=0$ .....(ii) Line (i) intersects the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ at Q. ∴ $\begin{vmatrix} 3-1 & 8-3 & 2-2 \\ a & b & c \\ 2 & 4 & 3 \end{vmatrix}=0$ ⇒ $\begin{vmatrix} 2 & 5 & 0 \\ a & b & c \\ 2 & 4 & 3 \end{vmatrix}=0$ ⇒ $2(3b-4c)-5(3a-2c)=0$ ⇒ $15a-6b-2c=0$ .......(iii) On solving Eqs. (ii) and (iii) by cross-multiplication, we get $\frac{a}{-4-12}=\frac{b}{-30+6}=\frac{c}{-18-30}$ ⇒ $\frac{a}{-16}=\frac{b}{-24}=\frac{c}{-48}\Rightarrow\frac{a}{2}$ $=\frac{b}{3}=\frac{c}{6}$ Substituting a, b and c in Eq. (i), we get $\frac{x-3}{2}=\frac{y-8}{3}=\frac{z-2}{6}=\lambda$ [say] Let the coordinates of Q be $(2\lambda+3,3\lambda+8,6\lambda+2).$ It lies on the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ ∴ $\lambda+1=\frac{3\lambda+5}{4}=2\lambda$ ⇒ $\lambda=1$ ∴ Coordinate of Q = (2 + 3, 3 + 8, 6 + 2) i.e. Q =(5,11, 8). Hence, $PQ=\sqrt{(5-3)^2+(11-8)^2+(8-2)^2}$ $=\sqrt{4+9+36}=7$