Question

If $(x,y,z)\ne 0$ and $(\hat{i}+\hat{j}+3\hat{k})x+(3\hat{i}-3\hat{j}+\hat{k})y+(-4\hat{i}+5\hat{j})z=\alpha (x\hat{i}+y\hat{j}+z\hat{k}),$ then the number of possible different value(s) of $\alpha$ is

Answer 1

Given : $(\hat{i}+\hat{j}+3\hat{k})x+(3\hat{i}-3\hat{j}+\hat{k})y+(-4\hat{i}+5\hat{j})z=\alpha (x\hat{i}+y\hat{j}+z\hat{k})$
Equating the coefficient of $\hat{i},\hat{j}$ and $\hat{k},$ we get
$x+3y-4z=\alpha x\Rightarrow (1-\alpha )x+3y-4z=0\cdots \left(i\right)$
$x-3y+5z=\alpha y\Rightarrow x-(3+\alpha )y+5z=0\cdots \left(ii\right)$
$3x+y=\alpha z\Rightarrow 3x+y-\alpha z=0\cdots \left(iii\right)$
Since $(x,y,z)\ne 0$
Using $\left(i\right),\left(ii\right)$ and $\left(iii\right),$ we get
$\Rightarrow \left|\begin{array}{ccc}(1-\alpha )& 3& -4\\ 1& -(3+\alpha )& 5\\ 3& 1& -\alpha \end{array}\right|=0$
$\Rightarrow (1-\alpha )\left(\alpha \right(3+\alpha )-5)-3(-\alpha -15)-4(1+3(3+\alpha \left)\right)=0$
Simplifying, we get
$\Rightarrow -{\alpha }^3-2{\alpha }^2-\alpha =0$
$\Rightarrow -\alpha ({\alpha }^2+2\alpha +1)=0$
$\therefore \alpha =-1,0$