Let $\mathrm{A}(-1,1)$ and $\mathrm{B}(2,3)$ be two points and $\mathrm{P}$ be a variable point above the line $A B$ such that the area of $\triangle \mathrm{PAB}$ is $10$ . If the locus of $\mathrm{P}$ is $\mathrm{ax}+\mathrm{by}=15$, then $5 a+2 b$ is :

$\left| {\,\begin{array}{*{20}{c}}{13}&{16}&{19}\\{14}&{17}&{20}\\{15}&{18}&{21}\end{array}\,} \right| = $

Suppose $D = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ and $D' = \left| {\,\begin{array}{*{20}{c}}{{a_1} + p{b_1}}&{{b_1} + q{c_1}}&{{c_1} + r{a_1}}\\{{a_2} + p{b_2}}&{{b_2} + q{c_2}}&{{c_2} + r{a_2}}\\{{a_3} + p{b_3}}&{{b_3} + q{c_3}}&{{c_3} + r{a_3}}\end{array}\,} \right|$, then

The number of integers $x$ satisfying $-3 x^4+\operatorname{det}\left[\begin{array}{ccc}1 & x & x^2 \\ 1 & x^2 & x^4 \\ 1 & x^3 & x^6\end{array}\right]=0$ is equal to

Show that points $A(a, b+c), B(b, c+a), C(c, a+b)$ are collinear

Find the equation of the line joining $\mathrm{A}(1,3)$ and $\mathrm{B}(0,0)$ using determinants and find $\mathrm{k}$ if $\mathrm{D}(\mathrm{k}, 0)$ is a point such that area of triangle $\mathrm{ABD}$ is $3 \,\mathrm{sq}$ $\mathrm{units}$.