$\vec{\beta}=2 \hat{i}-\hat{j}+3 \hat{k}$
$\vec{\beta}=\vec{\beta}_{1}-\vec{\beta}_{2}$
$\overrightarrow {{\beta _1}} = \lambda (3\hat i + \hat j),\overrightarrow {{\beta _2}} = \lambda (3\hat i + \hat j) - 2\hat i + \hat j - 3\hat k$
$\vec{\beta}_{2} \cdot \vec{\alpha}=0$
$(3 \lambda-2) \cdot 3+(\lambda+1)=0$
$9 \lambda-6+\lambda+1=0$
$\lambda=\frac{1}{2}$
$\Rightarrow \vec{\beta}_{1}=\frac{3}{2} \hat{i}+\frac{1}{2} \hat{j}$
$\Rightarrow \vec{\beta}_{2}=-\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k}$
${\rm{ Now, }}{\vec \beta _1} \times {\vec \beta _2} = \left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
{\frac{3}{2}}&{\frac{1}{2}}&0\\
{ - \frac{1}{2}}&{\frac{3}{2}}&{ - 3}
\end{array}} \right|$
$=\hat{i}\left(-\frac{3}{2}-0\right)-\hat{j}\left(-\frac{9}{2}-0\right)+\hat{k}\left(\frac{9}{4}+\frac{1}{4}\right)$
$=\frac{3}{2} \hat{i}+\frac{9}{2} \hat{j}+\frac{5}{2} \hat{k}$
$=\frac{1}{2}(-3 \hat{i}+9 \hat{j}+5 \hat{k})$
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where $[x]$ denotes the greatest integer less than or equal to $x$. Let $f \circ:(-1,1) \rightarrow R$ be the composite function defined by $(f \circ g)(x)=f(g(x))$. Suppose $c$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is NOT continuous, and suppose $d$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is $NOT$ differentiable. Then the value of $c+d$ is. . . . .