Are Shiksha's NCERT Solutions for Vector Algebra updated as per the latest CBSE syllabus?

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ <!-- [if gte mso 9]><xml> </xml><![endif]--> are non-zero and $\overrightarrow{b}$ <!-- [if gte mso 9]><xml> </xml><![endif]--> and $\overrightarrow{c}$ <!-- [if gte mso 9]><xml> </xml><![endif]--> are non-collinear. $\overrightarrow{a}+5\overrightarrow{b}$ <!-- [if gte mso 9]><xml> </xml><![endif]--> is collinear with $\overrightarrow{c}$ <!-- [if gte mso 9]><xml> </xml><![endif]--> and $\overrightarrow{b}+6\overrightarrow{c}$ <!-- [if gte mso 9]><xml> </xml><![endif]--> is collinear with $\overrightarrow{a}$ <!-- [if gte mso 9]><xml> </xml><![endif]-->. If $\overrightarrow{a}+\alpha \overrightarrow{b}+\beta \overrightarrow{b}=0$ <!-- [if gte mso 9]><xml> </xml><![endif]-->, then find α + β.

Let $\overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k}and\overrightarrow{c}$ be a vector such that $\overrightarrow{a}+\left(\overrightarrow{b}\times \overrightarrow{c}\right)=\overrightarrow{0}and\overrightarrow{b}.\overrightarrow{c}=5.$ Then, the value of $3\left(\overrightarrow{c}.\overrightarrow{a}\right)$  is equal to……….

Let A, B, C be three points whose position vectors respectively are

  $\overrightarrow{a}=\widehat{i}+4\widehat{j}+3\widehat{k}$

$\overrightarrow{b}=2\widehat{i}+\alpha \widehat{j}+4\widehat{k},\alpha \in R$

$\overrightarrow{c}=3\widehat{i}-2\widehat{j}+5\widehat{k}$

If a is the smallest positive integer for which $\overrightarrow{a},\overleftarrow{b},\overrightarrow{c}$ are noncollinear, then the length of the median, in $\Delta ABC,$  through A is: