Question

x + y + z + 1 = 0
ax + by + cz + d = 0
a²x + b²y + x²z + d² = 0

Answer 1

$$\begin{gathered} \mathrm{These}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as} \\ x+y+z=-1 \\ ax+by+cz=-d \\ {a}^{2}x+b^{2}y+x^{2}z=-d^2 \\ D=\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right| \\ =\left|\begin{array}{ccc}1& 0& 0\\ a& a-b& b-c\\ a^2& a^2-b^2& b^2-c^2\end{array}\right|\left[\mathrm{Applying}{C}_2\to C_1-C_2,C_3\to C_2-C_3\right] \\ \mathrm{Taking}(b-a)\mathrm{and}(c-a)\mathrm{common}\mathrm{from}{C}_1\mathrm{and}{C}_2,\mathrm{respectively},\mathrm{we}\mathrm{get} \\ =(a-b)(b-c)\left|\begin{array}{ccc}1& 0& 0\\ a& 1& 1\\ a^2& a+b& b+c\end{array}\right| \\ =(a-b)(b-c)(c-a)\dots \left(1\right) \\ {D}_1=\left|\begin{array}{ccc}-1& 1& 1\\ -d& b& c\\ -d^2& b^2& c^2\end{array}\right|=-\left|\begin{array}{ccc}1& 1& 1\\ d& b& c\\ d^2& b^2& c^2\end{array}\right| \\ {D}_1=-(d-b)(b-c)(c-d)\left[\mathrm{Replacing}a\mathrm{by}d\mathrm{in}\mathrm{eq}.\left(1\right)\right] \\ {D}_2=\left|\begin{array}{ccc}1& -1& 1\\ a& -d& c\\ a^2& -d^2& c^2\end{array}\right|=-\left|\begin{array}{ccc}1& 1& 1\\ a& d& c\\ a^2& d^2& c^2\end{array}\right| \\ {D}_2=-(a-d)(d-c)(c-a)\left[\mathrm{Replacing}b\mathrm{by}d\mathrm{in}\mathrm{eq}.\left(1\right)\right] \\ {D}_3=\left|\begin{array}{ccc}1& 1& -1\\ a& b& -d\\ a^2& b^2& -d^2\end{array}\right|=-\left|\begin{array}{ccc}1& 1& 1\\ a& b& d\\ a^2& b^2& d^2\end{array}\right| \\ {D}_3=-(a-b)(b-d)(d-a)\left[\mathrm{Replacing}c\mathrm{by}d\mathrm{in}\mathrm{eq}.\left(1\right)\right] \\ \mathrm{Thus}, \\ x=\frac{D_1}{D}=-\frac{(d-b)(b-c)(c-d)}{(a-b)(b-c)(c-a)} \\ y=\frac{D_2}{D}=-\frac{(a-d)(d-c)(c-a)}{(a-b)(b-c)(c-a)} \\ z=\frac{D_3}{D}=-\frac{(a-b)(b-d)(d-a)}{(a-b)(b-c)(c-a)} \end{gathered}$$