How much will the variance vary?

So, lets derive a general formula:-
Let $x_i = x_1, x_2, x_3, \cdots$ be $n$ values of variable $X$ .
Let $k_i = k_1, k_2, k_3, \cdots$ be $n$ values of variable $K$ such that $k_i = yx_i$ where $y$ is any non-zero real number.
If $\overline{K}$ be the mean of the obversations of $k_i$ and $\overline{X}$ be the mean of the observations of $x_i$ .

$\begin{aligned} \overline{K} & = \dfrac{1}{n} \displaystyle \sum_{i = 1}^{n} k_i\\ \\ & = \dfrac{1}{n} \displaystyle \sum_{i = 1}^{n} yx_i\\ \\ & = \dfrac{1}{n} × \displaystyle \sum_{i = 1}^{n} x_i × ny\\ \\ & = y\left(\displaystyle \sum_{i = 1}^{n} x_i\right)\\ \\ \overline{K} & = y\overline{X}\\ \\ k_i - \overline{K} & = yx_i - y\overline{X} \color{#EC7300}{\left(\text{because} \, k_i = yx_i\right)}\\ \\ k_i - \overline{K} & = y\left(x_i - x\right)\\ \\ {\left(k_i - \overline{K}\right)}^2 & = y^2{\left(x_i - \overline{X}\right)}^2 \color{#EC7300}{\left(\text{Squaring on both sides}\right)}\\ \\ \displaystyle \sum_{i = 1}^n {\left(k_i - \overline{K}\right)}^2 & = y^2 \displaystyle \sum_{i= 1}^{n} {\left(x_i - \overline{X}\right)}^2\\ \\ \dfrac{1}{n} \displaystyle \sum_{i = 1}^n {\left(k_i - \overline{K}\right)}^2 & = \dfrac{1}{n} y^2 \displaystyle \sum_{i = 1}^{n} {\left(x_i - \overline{X}\right)}^2\\ \\ \dfrac{1}{n} \displaystyle \sum_{i = 1}^n {\left(k_i - \overline{K}\right)}^2 & = y^2 \left(\dfrac{1}{n} \displaystyle \sum_{i = 1}^{n} {\left(x_i - \overline{X}\right)}^2\right)\\ \color{#20A900}{\text{Var(K)}} & = \color{#20A900}{y^2 \text{Var(X)}} \end{aligned}$

So, this means if we multiply any set of observations by any non-zero real number $y$ , the variance increases $y^2$ times.

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So, in the given question observations are multiplied by $3$ .
So, new variance will be:-
$\begin{aligned} 11 × 3^2 & = 11 × 9\\ & = \color{#3D99F6}{\boxed{99}} \end{aligned}$

We know that because variance = (standard deviation)^2, the old SD = sqrt(11). When each element of the old set is multiplied by 3, the SD of the resulting set is also multiplied by 3, giving the new SD to be 3sqrt(11). Thus, the new variance is (3sqrt(11))^2 = 99.