(a)OR(b)

Find the equations of the diagonals of the parallelogram PQRS whose vertices are $\displaystyle \mathrm{P}(4,2,-6), \mathrm{Q}(5,-3,1), \mathrm{R}(12,4,5)$ and $\displaystyle \mathrm{S}(11,9,-2)$. Use these equations to find the point of intersection of diagonals.

Equation of diagonal PR: $\displaystyle \frac{x-4}{8}=\frac{y-2}{2}=\frac{z+6}{11}$ Equation of diagonal QS: $\displaystyle \frac{x-5}{6}=\frac{y+3}{12}=\frac{z-1}{-3}$ General points on $\displaystyle \mathrm{PR} \& \mathrm{QS}$ are $\displaystyle (8 \mathrm{k}+4,2 \mathrm{k}+2,11 \mathrm{k}-6)$ and $\displaystyle (6 \mathrm{t}+5,12 \mathrm{t}-3,-3 \mathrm{t}+1)$ for real numbers ' $\displaystyle \mathrm{k}$ ' and ' $\displaystyle \mathrm{t}$ ' respectively. For point of intersection of PR and $\displaystyle \mathrm{QS}: 8 \mathrm{k}+4=6 \mathrm{t}+5,2 \mathrm{k}+2=12 \mathrm{t}-3$ Solving, we get $\displaystyle \mathrm{k}=\frac{1}{2}, \mathrm{t}=\frac{1}{2} \therefore$ The point of intersection is $\displaystyle \left(8,3,-\frac{1}{2}\right)$ Or (b) Let direction ratios of the required line be $\displaystyle \mathbf{a , b , c}$ Since it is perpendicular to the two given lines, $\displaystyle a+2 b+3 c=0 ;-3 a+2 b+5 c=0$ Solving together, $\displaystyle a=4 \mathrm{k}, b=-14 \mathrm{k}, c=8 \mathrm{k}$ ∴ Equation of line is: $\displaystyle \frac{x+1}{4 \mathrm{k}}=\frac{y-3}{-14 \mathrm{k}}=\frac{z+2}{8 \mathrm{k}} \Rightarrow \frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$ Vector equation: $\displaystyle \overrightarrow{\mathbf{r}}=-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})$