Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}$. If $\vec{d}$ is the unit vector in the direction of $\vec{b}+\vec{c}$ such that $\vec{a} \cdot \vec{d}=1$, then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to

<p>$$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}, \vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in R$$<?p> <p>$\text { also, } \vec{b}+\vec{c}=(x+2) \hat{i}+6 \hat{j}-2 \hat{k}$</p> <p>$\vec{d} \text { is the unit vector in the direction of } \vec{b}+\vec{c}$</p> <p>$$\begin{aligned} & |\vec{b}+\vec{c}|=\sqrt{(x+2)^2+6^2+2^2} \\ & =\sqrt{40+(x+2)^2} \\ & \vec{d}=\frac{x+2}{\sqrt{40+(x+2)^2}} \hat{i}+\frac{6}{\sqrt{40+(x+2)^2}} \hat{j} -\frac{2}{\sqrt{40+(x+2)^2}} \hat{k} \\ \end{aligned}$$</p> <p>$$\begin{aligned} & \vec{a} \cdot \vec{d}=1 \\ & \frac{x+2+6-2}{\sqrt{40+(x+2)^2}}=1 \\ & x+6=\sqrt{40+(x+2)^2} \end{aligned}$$</p> <p>$$\begin{aligned} &(x+6)^2=40+(x+2)^2 \\ & x^2+36+ 12 x=40+x^2+4+4 x \\ & 8 x=8 \\ & \Rightarrow x=1 \\ &(\vec{a} \times \vec{b}) \cdot \vec{c}=[\vec{a} \vec{b} \vec{c}] \\ &=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & -5 \\ 1 & 2 & 3 \end{array}\right] \\ &=1(12+10)-1(6+5)+1(4-4) \\ &=22-11=11 \end{aligned}$$</p>