Question
Let $\vec{\text{a}}=\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}}$ and $\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$Find a vector $\vec{\text{d}}$ which is perpendicular to both $\vec{\text{a}}$ and $\vec{\text{d}}$ and $\vec{\text{c}}.\vec{\text{d}}=15.$

Answer

Given
$\vec{\text{a}}=\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}}$
$\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$
Since d is perpendicular to both a and b, it is parallel to $\vec{\text{a}}\times\vec{\text{b}}.$
Suppose $\text{d}=\lambda\big(\vec{\text{a}}\times\vec{\text{b}}\big)$ for some scalar $\lambda.$
$\text{d}=\lambda\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&4&2\\3&-2&7 \end{vmatrix}$
$=\lambda\big[(28+4)\hat{\text{i}}-(7-6)\hat{\text{j}}+(-2-12)\hat{\text{k}}\big]$
$\vec{\text{c}}.\vec{\text{d}}=15$ (Given)
$\Rightarrow\big(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}\big).\lambda\big(32\hat{\text{i}}-\hat{\text{j}}-14\hat{\text{k}}\big)=15$
$\Rightarrow\lambda(64+1-56)=15$
$\Rightarrow\lambda=\frac{5}{3}$
$\therefore\vec{\text{d}}=\frac{5}{3}\big(32\hat{\text{i}}-\hat{\text{j}}-14\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{d}}=\frac{1}{3}\big(160\hat{\text{i}}-5\hat{\text{j}}-70\hat{\text{k}}\big)$
Disclaimer: The question should contain "Which is perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}$ " instead of "Which is perpendicular to both $\vec{\text{a}}$ and $\vec{\text{d}}$ "

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