If $A\left(2\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\right), B\left(\lambda \hat{\mathrm{i}}+5\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right), C\left(-4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right)$ and $D\left(-\hat{\mathrm{i}}-2\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)$ are four points in space such that $\overrightarrow{AB}=x\overrightarrow{AC}+y\overrightarrow{AD}$ for some real numbers $x\ne 0,y\ne 0$ then $17\left(\lambda +9\right)=$

Given,

$A\left(2\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\right), B\left(\lambda \hat{\mathrm{i}}+5\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right), C\left(-4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right)$ and $D\left(-\hat{\mathrm{i}}-2\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)$ are four points in space such that $\overrightarrow{AB}=x\overrightarrow{AC}+y\overrightarrow{AD}$

So, $\overrightarrow{AB}=x\overrightarrow{AC}+y\overrightarrow{AD}$

$\Rightarrow \left(\lambda -2\right)\hat{\mathrm{i}}+4\hat{\mathrm{j}}+5\hat{\mathrm{k}}=x\left(-6\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)+y\left(-3\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)$

Now on comparing coefficient $\hat{\mathrm{i}}, \hat{\mathrm{j}} \& \hat{\mathrm{k}}$ we get,

$\Rightarrow \left(\lambda -2\right)=-6x-3y ........\left(1\right)$

$\Rightarrow 4=2x-3y ................\left(2\right)$

$\Rightarrow 5=3x+4y ...............\left(3\right)$

Now solving $\left(2\right) \& \left(3\right)$ we get,

$x=\frac{31}{17} \& y=\frac{-2}{17}$

Now putting the value of $x \& y$ in equation $\left(1\right)$ we get,

$\left(\lambda -2\right)=-6\times \frac{31}{17}-3\times \frac{\left(-2\right)}{17}$

$\Rightarrow \lambda =\frac{-146}{17}$

Then $17\left(\lambda +9\right)=17\left(\frac{-146}{17}+9\right)=7$